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GEOMETRICAL OPTICS 



GEOMETRICAL OPTICS 



BY 



ARCHIBALD STANLEY PERCIVAL 

M.A., M.B., B.C. CANTAB. 

SENIOR SURGEON NORTHUMBERLAND AND DURHAM EYE INFIRMARY 
AUTHOR OF "OPTICS," "PRACTICAL INTEGRATION," "PRESCRIBING OF SPECTACLES 1 




WITH 



LONGMANS, GREEN, AND CO, 

39 PATERNOSTER ROW, LONDON 
NEW YORK, BOMBAY, AND CALCUTTA 

1913 

[All rights re se 



PREFACE 

THIS book is primarily intended for medical students as a 
text-book on the subject of Geometrical Optics for their 
preliminary scientific examinations, though it practically 
contains all the Optics required by an ophthalmic surgeon. 
It is hoped that it will also prove of service to students of 
physics, as some knowledge of the subject is indispensable 
if the laboratory experiments are to be understood. 

It requires prolonged and deep study to form any vivid 
conception of the now accepted theory of light, and in all 
elementary books the form in which the undulatory theory 
is presented is so crude that it is both untrue and useless. 
The subject of Physical Optics therefore has been avoided 
entirely; indeed, I am convinced that no thorough elemen- 
tary knowledge of that intricate subject can be obtained in 
the short space of time allotted to the student for studying 
Optics. 

As an introduction to mathematical analysis the subject 
of Geometrical Optics has no equal, for it insists on the 
importance of paying due attention to the meaning of 
algebraic signs, and it is also an easy introduction to several 
somewhat difficult mathematical conceptions. For instance, 
the vectorial significance of the line BA being considered as 
equal to AB, or AB taken in the reverse direction, opens 
up a new vista to the student of Euclid and elementary 
geometrical methods: equally novel is the conception of a 
virtual image. At the same time every student can verify 
for himself the results of his calculations so simply by 
experiment that it will convince him of the reality of the 
analytical methods employed. 

263251 



vi PEEFACE 

I have embodied in the text all that can be reasonably 
demanded in any preliminary examination in science, while 
the Appendix contains matter that will be of service sub- 
sequently in a professional career. The starred paragraphs 
may be omitted at a first reading, as a knowledge of them 
would be rarely required in the examination-room. The 
subject of Cardinal Points is always neglected in the elemen- 
tary books ; it has here been treated, I venture to think, in 
a much simpler manner than is customary in the solution 
of the problem; for the ingenious graphic solution on 
p. 106 I am indebted to Professor Sampson. The subject is 
of first importance in understanding the optical properties 
of the eye, and it is of the greatest value in dealing simply 
and readily with many otherwise difficult questions. 

The reader will find a list of all the important formulae 
for ready reference at the end of the book, and he will notice 
that some of them are of universal application, and that by a 
simple transformation, formulae for refraction can be converted 
into the corresponding formulae for reflection. The proofs 
while preserving their rigid character are made as simple as 
possible, and the utmost care has been taken to include only 
what is of practical application, excluding all that is of 
merely academic interest. 

ARCHIBALD STANLEY PERCIVAL. 

17, CLAREMONT PLACE, 
NEWCASTLE-UPON-TYNE, 
Dec. 1912. 



CONTENTS 



PAGE 

I. ILLUMINATION PINHOLES SHADOWS 1 

IL REFLECTION AT PLANE SURFACES 11 

III. REFLECTION AT A SPHEBICAL SURFACE 18 

IV. REFRACTION AT PLANE SURFACES 34 

V. REFRACTION AT A SPHERICAL SURFACE 49 

VI. LENSES 68 

APPENDIX 109 

ANSWERS 129 

INDEX 131 



CHAPTEE I 

ILLUMINATION PINHOLES SHADOWS 

WE shall here confine ourselves to the study of some of the 
simplest properties and laws of Light. When we say that 
we see the sun or a tree we merely mean that we see the 
light that comes from them ; the sun, of course, sends out 
light of its own, whereas the tree passes on or reflects the 
light chat it receives from something else. It is obvious 
that in neither case do we see the thing itself ; we are only 
conscious of a certain sense-impression derived from it. The 
nature of this sense-impression and the way in which it is 
developed from a physical stimulation of the retina, are 
problems that are still engaging the attention of physio- 
logists and psychologists ; with questions such as these the 
science of Optics does not deal. 

We shall commence our study by the consideration of the 
following laws : 

LAW 1. In a homogeneous medium light is propagated in 
every direction in straight lines. 

LAW 2. The intensity of illumination varies inversely as 
the square of the distance from the source of 
light, and it is greatest when the angle of 
incidence is 0. 

LAW 3. When the incident rays of light are parallel, the 
intensity of illumination varies as the cosine 
of the angle of incidence. 

Our experience of shadows confirms the truth of Law 1. 
A luminous point in space sends out light in all directions, 



2 GEOMETEICAL OPTICS 

illuminating a rapidly enlarging sphere. It is the radii of 
this sphere that are called rays of light ; they are simply the 
directions in which the light is travelling, and when this 
direction is altered by reflection or refraction the rays con- 
sidered are bent at an angle. But for either reflection or 
refraction to occur a heterogeneous medium must be 
encountered; so that we conclude that in a homogeneous 
medium light is propagated in every direction in straight 
lines. 

Experiments with diffraction gratings show us that Law 1 is not 
strictly true, and, had we to treat of the phenomena of Diffraction and 
Polarization, we should have to explain in detail the electromagnetic 
theory of light, or at any rate to give an account of some form of wave 
theory. As these subjects do not now concern us, we may regard the 
directions in which light travels as straight lines or rays, and in this 
way the elementary study of optics will be much simplified. 

The second and third laws will require some careful 
explanation and consideration before they will be accepted 
by the reader. 

Illumination. Let S be a source of light, and consider 
the light that is being propagated from S in the direction of 
the screens AK and BL (Fig. 1). It will be noticed that, 



FIG. i. 



since light travels in straight lines, the amount of light (Q) 
that falls upon the square AK is the same as that which 
would illuminate the larger square BL if AK were removed. 
Now, in the diagram SB is made equal to 3 SA, so the area 
of the square BL is nine times the area of AK. Hence the 



ILLUMINATION 3 

illumination per unit area, or the intensity of illumination, of 
BL is of that of AK. 

Indeed, B T : A= 1:9 " 

It is obvious, then, that the intensity of illumination 
varies inversely as the square of the distance from the source 
of light. 

Nearly all photometers are based upon this law ; we will 
describe that designed by Eumford. Suppose that we wish 
to compare the intensities, I and L, of illumination of two 
lamps ; these are placed at distances d and D from a white 
screen so that two sharp shadows of a vertical rod placed a 
few inches in front of the screen are thrown upon it ; each 
shadow is then illuminated by only one of the lamps. Thus 
if L, the standard light, be regarded as 1, the other lamp is 
moved backwards or forwards until the position d is found 
in which both shadows appear equally dark. Each lamp is 
then sending an equal amount of light to the screen, and 
the relative illuminating power is given by the ratio of 
the squares of their respective distances (d? and D 2 ) from 
the screen. The light that the two shadows receive are 

respectively -^ and^; consequently when these are equal, 

1 : L = d 2 : D 2 . Thus if D the distance of the standard 
light be 2 feet, while d = 4 feet, 

1 : 1 = d? : D 2 = 16 : 4 = 4 : 1 

or I is four times the intensity of the standard light. 

It is commonly said that four candles at a distance of 

2 feet give the same illumination as one candle at a 
distance of 1 foot ; this, although true, could not be satis- 
factorily proved by a photometer which measures only the 
intensity of illumination from a point source of light. 

It may be noted that the accuracy with which such tests 
can be made depends upon the " Light Sense " of the observer, 
so that if the values of I and L be known, this photometer 
may be used as a test of the observer's " Light Difference." 

On referring to Fig. 1, it will be noticed that both AK 



4 GEOMETKICAL OPTICS 

and BL are at right angles to SB. Now, the angle of incidence 
(0) is the angle which a line at right angles to the surface 
of the screen makes with the incident ray ; in this case, there- 
fore, the angle of incidence is 0. If the screen were inclined 
either forwards or backwards, turning round its base line, it 
is clear that the intensity of illumination per unit area of 
the screen would be diminished, for in either case some of 
the cone of light that is represented would not fall upon the 
tilted screen. The diminution of illumination approximately 
follows the cosine law (see below) when the incident cone 
of light is not too divergent. 

If the illuminant were a beam of parallel rays, or were at 
an infinite distance as compared with the distance AB, the 
law of inverse squares ceases to have any intelligible meaning ; 
for then both SA and SB are infinite, and the illumination of 
a surface exposed to such a light is constant whatever its 
distance. For instance, the light from a morning sun on a 
screen is practically the same as the light on a similar screen 
placed at a mile's distance to the west. 

If I be the intensity of illumination a screen receives from 
a beam of parallel rays when the angle of incidence is 0, it 
is clear that when the screen is tilted so that the angle of 
incidence is 0, the illumination is I cos 0. 

Apparent Brightness. The distinction between brightness 
and illumination is hardly made clear enough in the books. 
Illumination refers to the physical condition of the object 
when illuminated, whereas brightness refers to the resulting 
sense-impression produced in the observer. It will clearly 
depend upon his physiological condition ; but, neglecting this 
for the moment, let us consider how an illuminated screen at 
BL will appear to an observer at S; 

The brightness of an object is naturally measured by the 
amount of light it sends to the eye per unit area of its 
apparent size; in other words, the brightness (B) of an 
object is directly proportional to the quantity of light (Q) 
that it sends to the pupil, and inversely proportional to the 

apparent area (A) of the surface observed, or B = ~. 

A 



APPARENT BRIGHTNESS 5 

Since, however, both Q and A are each of them inversely 
proportional to the square of the distance of the luminous 
object, the apparent brightness is independent both of its 
size and of its distance, presuming of course that the medium 
is clear and no adventitious absorption occurs in it. If the 
physiological conditions be the same, the apparent brightness 
of an object merely depends upon the intrinsic luminosity of 
the object. A red-hot iron ball is indistinguishable from a 
circular disc of iron at the same temperature, showing that 
the brightness is independent of the inclination of the 
periphery of the ball to the line of sight. The sun and 
moon may both be regarded as approximately spherical, yet 
both appear to the naked eye as flat discs of uniform bright- 
ness. There are, however, several physiological conditions 
that greatly affect the apparent brightness of luminous 
bodies. 

(1) The size of the pupil. The brightness (B) must vary 
as the area of the pupil. The brightness of an object will 
be much diminished by viewing it through a small pinhole, 
for in that case less light enters the eye, as the size of the 
pupil is virtually diminished. 

(2) The condition of retinal adaptation. Prolonged stay 
in a dark room will much increase the apparent brightness 
beyond what is due to the consequent dilatation of the pupil. 

When high powers of a microscope or a telescope are 
used, the brightness of the image is much diminished, as 
then only part of the pupil is filled with light. This is 
practically equivalent to lessening the area of the pupil (1). 
It is impossible by any optical arrangement to obtain an 
image whose brightest part shall exceed the brightest part 
of the object. If allowance be made for the loss of a certain 
amount (about 15 %) of light by reflection and imperfect 
transparency of the lenses, the brightness of the image is 
equal to the brightness of the object. 

There is one case in which this law does not hold good, 
and this also depends upon physiological reasons. If the 
object be extremely small, it may yet if excessively bright 
succeed in stimulating a retinal cone, and so cause a visual 



6 GEOMETKICAL OPTICS 

impression of a tiny point of light, even although its image 
does not cover the whole surface of the cone. If such an 
object be magnified until its image covers the cone, there will 
be no increase of its apparent size although an increased 
amount of light will be entering the eye. When, for 
instance, stars are observed through a telescope, their 
apparent size is not increased, for their image does not 
extend beyond one retinal cone, but all the light that falls on 
the object-glass may by a suitable eyepiece be concentrated 
on the observer's pupil, except that lost by transmission 
through the lenses. If, then, a denote the fraction of the 
incident light that is transmitted through the telescope 
(about 0*85), and O denote the area of the object-glass, and e 

that of the pupil, the increase of brightness will be a . Or 

Q 

if o and p be the diameters of the object-glass and the pupil 

o 

respectively, the increase of brightness will be a -3 , for the 

areas of circles are proportional to the squares of their 
diameters. If the pupil be regarded as of unit diameter we 
get the expression ad 2 . This is what astronomers call the 
" space penetrating power " of a telescope, that is its power of 
rendering very small stars visible. 

Precisely the same conditions obtain with the ultrami- 
croscope, which is a device for bringing into view fine 
granules that are smaller than the limit of the resolving 
power of the instrument. By means of dark-ground illumi- 
nation these fine particles are relatively so brightly lighted 
that they succeed in stimulating a retinal cone even though 
their image does not cover its whole surface. ^Resolution 
does not occur; whatever its real shape the granule will 
appear as a circular point of light, so that all one can really 
say is that something very small is present. 

Visual Angle. On referring again to Fig. 1 it will be 
seen that if the observer's eye were situated at S the small 
square AK would completely obliterate the larger square BL, 
as they both subtend the same angle at S. Such an angle is 
called the visual angle, and it is obvious that if one does not 



PINHOLES 7 

know the distance of an object one can form no real idea of 
its size. The expression " apparent size " is quite illusory ; it 
conveys no clear meaning. For instance, some will say that 
the sun's apparent size is that of a dinner plate, others that 
of a saucer, but it will be found on trial that a threepenny 
bit at a distance of 6 feet will completely obliterate it, yet 
few would assert that its apparent size was that of a three- 
penny bit even when held at arm's length. The importance 
of the visual angle will appear when we deal with magnifi- 
cation. 

Pinholes. If a pinhole be made in a card, and this be 
held between a candle and a screen, an inverted image of the 
candle flame will be formed upon the screen. The nearer 
the screen is brought to the pinhole the smaller and sharper 
will be the image. If the candle be brought nearer the pin- 
hole, the image will be larger but less sharp. Finally, if the 
hole in the card be made larger the image will appear 
brighter, but its definition will be again diminished. 

The explanation is simple, as will appear from an exami- 
nation of Fig. 2. Every point of the candle flame is sending 
out light in all directions, and all that falls upon the card is 
intercepted by it, so that its shadow is thrown upon the 
screen. From each point of the candle flame, however, there 
will be one tiny cone of light that will make its way through 
the pinhole. On the screen the section of this cone will 
appear as a small patch of light. Thus the point A of the 
flame will be represented by the patch a, and the point B by 
the patch & on the screen. Clearly the height ab of this 
inverted image will be proportional to the distance of the 
screen from the pinhole ; if AB be brought nearer the card 
each pencil will form a more divergent cone, or if the hole be 
made larger the same result will occur ; hence in both these 
cases the patches a and & will be larger and the definition 
will be impaired. 

If care be taken that these patches of light are not too 
large, an image sufficiently sharp for photographic purposes 
may be obtained. For some purposes a pinhole camera made 
out of a preserved meat tin may prove a more satisfactory 



8 



GEOMETRICAL OPTICS 



instrument than one provided with a lens. The disad- 
vantage of a pinhole camera is of course the feeble illumi- 
nation of the image, which makes a very long exposure 
necessary. 




FIG. 2. 



The best size of pinholes for photographic work, according to 
Abney, is determined by the approximate formula 

p = 0-008 Jd 

Thus, if the distance between the pinhole and the plate be 9 ins., the 
diameter of the pinhole should be O'OOS x 3 = 0'024 in. 

Shadows. When the source of illumination may be 
regarded as a point, it is clear that an opaque obstacle 
like AK in Fig. 1 will be lighted only on its proximal sur- 



SHADOWS 



9 



face, while the prolongation of the cone of light indicates the 
cone of shadow cast by AK. 

If, however, the source of light be a luminous body 
possessing innumerable points, the object will be illuminated 
by innumerable cones of light, and we must imagine a shadow 
for each of them behind the object. The space behind the 
object which is common to all of these shadow cones will 
represent the area of total shadow, or umbra. There will, 
however, be a space outside this which is only in shadow as 
regards part of the luminous body while it receives light from 
another part of it, and is consequently partially illuminated. 
This is the area of part-shadow, or penumbra. In the adjoin- 
ing illustration (Fig. 3) two opaque bodies are represented, 




FIG. 3. 

one being smaller and the other larger than the luminous 
body which is placed between them. In each case for the 
sake of clearness the limits between umbra and penumbra 
have been more sharply defined than they should be. When 
a total eclipse of the sun occurs, the moon is so situated that 
it intercepts all the light coming from the sun towards the 
earth ; the earth is then in the umtoa : when a partial eclipse 
occurs, the earth is in the penumbra. 



10 GEOMETRICAL OPTICS 

QUESTIONS. 

(1) What is the height of a tower that casts a shadow 52 feet 6 
inches in length on the ground, the shadow of a stick 3 feet high being 
at the same time 3 feet 6 inches long ? 

(2) A pinhole camera, the length of which is 7 inches, forms an 
inverted image 4 inches high of a house that is in reality 40 feet high. 
What is the distance of the camera from the house ? 

(3) A gas lamp distant 5 feet and an electric light distant 150 feet 
throw on an opposite wall two shadows of a neighbouring post. If 
these two shadows are of equal intensity, what is the relative illuminat- 
ing power of the lights ? 

(4) If the electric light in (3) were raised vertically to such a height 
that its distance from the wall were 300 feet, what would be approxi- 
mately the relative intensities of the shadows ? 



CHAPTER II 

REFLECTION AT PLANE SURFACES 

WE must now consider the manner in which light is reflected 
by polished surfaces. The following two laws when properly 
understood explain every possible case of reflection : 

LAW 1. The reflected ray lies in the plane of incidence. 

LAW 2. The angles which the normal (at the point of 
incidence) makes with the incident and with 
the reflected ray are numerically equal, but 
they are of opposite sign. 

These laws hold good whether the surface be plane or 
curved ; in the latter case we have only to draw a tangent 
at the point of incidence, and consider the ray reflected at 
this plane. The meaning of the technical terms and the 
sign of an angle will be explained in the next paragraph. 

Reflection at Plane Surfaces. Let AB represent a plane 
reflecting surface (Fig. 4), and let S be a luminous point 
sending out light in all directions. Now, if SN be one of 
these directions, SN represents an incident ray, N the point 
of incidence, and NY the normal at that point. Now, the 
plane of incidence is that plane that contains both the 
incident ray and the normal, so the plane of the paper is the 
plane of incidence. According to the laws just stated NQ is 
the corresponding reflected ray, for the angle of incidence 
YNS is numerically equal to the angle of refraction YNQ, 
which lies in the plane of the paper, and is of opposite sign to 
YNS. The angle YNS is measured by the rotation at N of 
a line in the direction NY to the position NS, i.e. in the 
direction of the hands of a clock, whereas YNQ is measured 

11 



12 



GEOMETEIOAL OPTICS 



in the counter-clockwise direction. This is what is meant by 
the sign of an angle, so that if YNQ be denoted by $' and 
YNS by 0, we have in reflection 0' = - <. 

If, then, there be an eye in the neighbourhood of Q it will 
receive light coming towards it in the direction NQ ; but we 
have not yet found from which point in this line it will appear 
to have come. To do this we shall have to take another 
incident ray, SM say, and discover where the corresponding 
reflected ray intersects the previous one. 




FIG. 4. 

It is clear that as the incident and reflected rays make 
equal angles with the normal they must make equal angles 
with the mirror ; so we made the angle BMR numerically 
equal to AMS, and produce the reflected ray RM to meet QN 
produced in S'. Then S' is the point from which the light will 
appear to have come. 

Now, it is clear that in the triangles SMN and S'MN we 
have the base MN common, and the angles at the base equal, 
so the triangles must be equal ; and if we join SS', we see that 
the two sides SM, MA are equal to the two sides S'M, MA, 
and the included angle SMA is equal to S'MA ; so S'A is 
equal to SA and the line SS' is at right angles to AB. 

Similarly, it may be shown that any other ray in the same 



CONSTRUCTION OF THE IMAGE 



13 



plane will be reflected in such a direction that when produced 
backwards it will meet SS' in S'. Now, if we suppose the 
paper to be revolved about an axis SS', the figure will repre- 
sent the course of incident and reflected rays in every plane. 
Hence all the rays that fall upon the mirror from S, whatever 
the plane of their incidence, will be so reflected that the 
prolongations of these reflected rays will intersect at the 
point S', so that an image of S will be formed at S'. 

It will be noted that this image has no real existence; the 
reflected rays do not really 
come from the back of the 
mirror, they only appear 
to come from the point S' : 
such an image is called a 
virtual image. Subse- 
quently we shall deal with 
real images, and the dis- 
tinction between them is 
simply this : Real images 
are formed "by the inter- 
sections of the reflected (or 
refracted) rays themselves ; 
virtual images are only 
formed by the intersections 
of their prolongations. 

Construction of the 
Image of an Object. Sup- 
pose that it is required 
to construct the image of the object Pp as seen in the 
mirror NM (Fig. 5); all we have to do is to draw lines 
perpendicular to the mirror (or its prolongation) from P 
and p to Q and q, so that Q and q are as far behind the 
mirror as P and p are in front of it ; then Qq is the virtual 
image of Pp. If an eye be situated in the neighbourhood of 
E, and we wish to show the actual course of the light that 
reaches the eye from P, we join EQ, cutting the mirror at M, 
and then we join MP. The light that forms the image Q for 
an eye at E reaches it by the path PM and ME, Another 




FIG. 5. 



14 GEOMETEICAL OPTICS 

eye at K' will see the same image Q by means of light that 
travels in the direction PN and NE'. 

It will be noted that the image Qq is similar and equal 
to the object "Pp, for they both are similarly situated with 
respect to the mirror. Since, however, the object and the 
image face each other, the observer gets a view of that side of 
the object that he could not see without turning it round. 
Hence he thinks the object has been turned round so that its 
right and left sides have been interchanged. For this reason 
the image of a printed letter will appear " perverted," that 
is, it will appear upright, but it will resemble the type from 
which it was printed. 

Deviation produced by Rotation of Mirror. If the 
mirror AB (Fig. 6) be slowly rotated into the position cib, 




the image S' will appear to rotate through an angle twice 
as great in the same direction, so that finally it will appear 
to have moved to S". The explanation is obvious. If S be 
the object, when the mirror is in the position AB its image 
will be at S' ; on rotating the mirror through an angle 9 its 
normal will also rotate through the same angle, so that 
SIY = 9, but then reflection will occur in the direction 
IK, and the image of S will appear at S". Clearly, 
SIS" = SIE = 29. 

There are several important applications of this principle 
in daily use ; as examples we may mention the sextant and 
the mirror galvanometer, but we will describe in greater 
detail the laryngoscope. When this instrument is used, the 
mirror, which is inclined at an angle of 45, is placed at the 



EEPEATED KEFLECTION 



15 



back of the pharynx; when properly placed an image of 
the larynx is seen with its axis horizontal. Since the image 
and object are similarly situated with respect to the mirror 
and not to the observer, the anterior parts of the larynx 
(epiglottis, etc.) are represented in the upper part of the 
image, while the posterior structures (arytenoids, etc.) occupy 
the lower portion of the image. 

Repeated Reflection at Inclined Mirrors. When an object 
is placed between two plane mirrors inclined at an angle, a 
limited number of images may be seen by an observer in a 




FIG. 7. 

suitable position. If a be the angle between the mirrors, 

,.-180 , 

and it = 7i, the number of images seen by an eye in a 

suitable position will be 2n - 1, if n be an integer. Moreover, 
the object together with its images then forms a perfectly 
symmetrical figure with respect to the reflecting surfaces. 
Fig. 7 represents two mirrors, AC and BC, inclined at an 
angle of 60 with an object PQ between them. As 2n 1 in 
this case is 5, 5 images are seen, and the symmetry of the 
figure is evident. This is the principle of the kaleidoscope. 



16 



GEOMETEICAL OPTICS 



1 80 
Suppose that a = 40, then n = -^- or 4 J ; in such a case 

it will depend entirely on the position of the observer 
whether 4 or 5 images are seen. 

The position of the inages is easily found by the construction now 
familiar to the reader. Let AC, BC (Fig. 8) be two plane mirrors 
inclined at an angle of 90, and let^? denote the position of an object 
between them. From p draw perpendiculars to each mirror, and 
produce them to pf and p^ so that pf and p* are as far behind the 




FIG. 8 



mirrors AC and BC respectively asjp is in front of them. Then pf and 
pj are two of the images. To find the position of the third image p n , 
we must produce AC and BC and consider these prolongations as 
mirrors and the images we have already found as objects. That is 
to say, we make p u either at an equal distance behind AC produced 
to that of p^ in front of it, or we make p n an equal distance 
behind BC produced to that of pf in front of it. It will be noticed 
that whichever construction is used, p n occupies the same place. To 
find the path of light from p n to an eye at E, join jp n E, cutting BC 
in Q ; join pfQ, cutting AC in K ; join jpR. Then the path of the light 
from p is ^B, RQ, QE. If E had been close to AC, on joining _p n E 
the mirror AC instead of BC would have been cut, so that in such a 
case the last reflection would have been from the mirror AC, 



EEPEATED REFLECTION 17 

must be joined to the intersection on AC. These constructions are of 
little importance except for examination purposes. 

When the mirrors are parallel to each other, the angle 

1 80 
between them is 0, and since -^- = oo , an infinite number 

of images might be seen if certain physical conditions did 
not prevent their observation. 



QUESTIONS. 

(1) Two parallel mirrors face each other at a distance of 3 feet, 
and a small object is placed between them at a distance of 1 foot from 
one of them. Calculate the distances from this mirror of the three 
nearest images that are seen in it. 

(2) Two mirrors, AC and BC, are inclined at an angle of 45 ; an 
object, P, is so placed that ACP = 15. How many images will an 
observer halfway between the mirrors see ? Trace the path of light 
that gives rise to the second image p n b seen in BC. 

(3) If in the last example BC were rotated so that ACB = 135, 
how many images would an observer (E) halfway between the 
mirrors see ? In what position would he see more ? 



CHAPTEK III 

REFLECTION AT A SPHERICAL SURFACE 

EEFLECTION at irregularly curved surfaces can only be treated 
by the method mentioned on p. 11. At the point of 
incidence a tangent plane is drawn, and for that particular 
incident ray the reflection is considered as taking place at 
that tangent plane. Fortunately, however, there are much 
less tedious methods of dealing with reflection at spherical 
surfaces, with which alone we are now concerned ; these we 
proceed to describe in their simplest form. 

Concave Spherical Mirrors ; Axial or Centric Pencils. Let 
AK represent a concave mirror, its surface being part of a 
sphere of which the centre is at C (Fig. 9). Any line drawn 
through C to the mirror is called an axis of the mirror, and 
when the parts of the mirror on either side of the axis are 
symmetrical it is called the Principal Axis. In the figure 
CA is the principal axis, and A the vertex of the mirror. 
Let P be a luminous point on the axis distant PA or p from 
the mirror, and consider the incident ray PK when the point 
K is near the vertex A. Join CK and make the angle CKQ 
numerically equal to the angle of incidence CKP. If now 
the figure be supposed to revolve round the principal axis 
PA, it is clear that PK will trace out the limits of a thin 
axial cone of incident light, while the point K will trace 
out a narrow circular zone on the mirror, and the reflected 
rays from this zone must all intersect the axis at the point 
Q. Consequently Q must be the image of P as formed by 
reflection at this narrow zone. Let CA, the radius of the 
mirror, be denoted by r, and let QA be denoted by q. 

18 



CONCAVE SPHERICAL MIREOKS 



19 



Now, since in the triangle PKQ the vertical angle is 
bisected by KC, which cuts the base at C 



(Euc. vi. 3) 



QK ~~ CQ ~" CA - QA r-q 



When K is very near the vertex A, PK and QK will be 
almost identical with PA and QA; under these circum- 




FIG. 9. 

stances we may replace PK and QK by p and q, and so we 
get the following formula for a very thin centric pencil : 

P = p-r 

.'. pr pq = pq rq or qr + pr = 2pq 
On dividing by pqr we obtain the formula 



pqr 

The point Q marks the situation of the real image of P ; 
it is a real image because it is formed by the intersection 
of the reflected rays themselves (see p. 13). Again, Q is 
often called the conjugate focus of P. It is termed a focus 
(fireplace) because the place where heat rays intersect is 
hotter than any other. If, for instance, the mirror be used 
to form an image of the sun, a piece of paper placed at the 



20 GEOMETKICAL OPTICS 

focus will be burnt up. The term "conjugate" implies that 
if the object P be placed at Q, the image will then be in the 
old situation of P. In fact, P and Q are mutually con- 
vertible whenever the image formed at Q is real. 

We will now consider the formula a little more closely. 
If PA or p be diminished until p = r y QA or q becomes 
greater until q becomes equal to r, for 



q r p r r r 

This only means that when the luminous point is placed 
at C all the incident light reaches the mirror in the direction 
of its radii, or the normals to the surface, so that the 
reflected light must travel back by the same path ; the image 
Q is then coincident with the object at C. 

If the object P be brought still closer to the mirror 
(between C and F), the image Q will be formed at a greater 
distance from it ; in fact, the situations of P and Q will be 
interchanged. 

Now let us suppose the object P to be removed to a great 
distance ; the image Q will be formed nearer the mirror. In 
the limiting case when the distance of P is infinite, let us 
see what our formula tells us. Here p = oo . 

so + - = -. But 1 = 0, so - = - 
cc q r oo q r 

This means that if the object is at an infinite distance, or, 
what comes to the same thing, if the incident beam consist 
of parallel rays, the corresponding focus is at a point F such 
that FA = CA. This point F is called the Principal Focus, 
and the distance FA is usually symbolized by/. Clearly, 
as the direction of light is reversible, it follows that if a 
luminous point be placed at F the reflected rays will form a 
beam of light parallel to the axis. 

T 

Since we now know that / = ~, we may give the previous 
formula in its more usual form 

M = U/ + /=i 

P 2 / P 2 



SIGNS 21 

What will happen if the object be placed nearer the 
mirror than F? Our formula will tell us. Suppose that 
/ = 4, and that p = 3 ; 



QA or q is then 12. What is the meaning of the 
negative sign ? We have regarded the direction from left to 
right (e.g. PA and CA) as positive ; the opposite direction, 
from right to left, must therefore be the negative direction. 
Consequently, QA must now be measured in this direction, 
that is, the point Q must be behind the mirror. The reader 
is urged to draw for himself a diagram illustrating this case ; 
that is, let him draw an arc of 8 cm. radius with its focus at 
4 cm. from the mirror, and from a point P, 3 cm. from the 
mirror, let him draw any incident ray, PK, to the mirror ; 
let him then mark a point Q, 12 cm. to the right of (behind) 
the mirror, and join QK and produce it. He will find that 
the normal CK bisects the angle between PK and the pro- 
longation of QK, showing that KQ is the prolongation of the 
reflected ray in this case in other words, Q is a virtual 
image of P, as it is formed, not by the reflected ray itself, but 
by its prolongation (see p. 13). 

Whenever Q is a virtual image, P and Q are no longer 
interchangeable ; in this case, for instance, the object if placed 
in the situation of Q would be behind the reflecting surface 
of the mirror, so no image would be formed. 

Signs. This case will indicate the importance to be 
attached to the meaning of algebraic signs. It will be found, 
if due attention be paid to them when thin axial pencils 
are being considered, that there are only two formulae that 
need be remembered for reflection or refraction at single 
spherical surfaces, and for lenses of any kind or for any com- 
bination of them. The essential thing is to be consistent 
during any calculation ; any inconsistency may lead to totally 
erroneous results. 

In all optical problems it is most important to remember 



22 GEOMETEICAL OPTICS 

that the symbol PC for a line, such as that in Figs. 9 and 10, 
denotes the distance from P to C, so that it not only 
expresses a length, but also the direction in which that 
length is measured. Euclid uses the term PC as identical 
with CP. We, however, must regard PC as a vectorial 
symbol, and therefore equal to CP. Consequently 

PC = PA + AC = PA - CA = AC - AP 

Similar vectorial expressions will be used throughout this 
book. 

We have adopted the usual conventions that directions 
from left to right are considered positive, and those from 
right to left negative. Further, we shall regard directions 
from below upwards as positive, and those from above down- 
wards as negative. As regards angles, we shall designate 
the directions of rotation in the usual way. All rotations in 
the counter-clockwise direction are considered positive, while 
all clockwise rotations are considered negative. The term 
" Standard Notation " will in this book be used to denote 
this device of signs to indicate these various directions. 

In optics more blunders are due to the neglect of the 
meaning of signs than to any other cause, so it is well worth 
while devoting some attention to the subject. When a cor- 
rect mathematical formula is given, one knows that it must 
be universally true, whatever values and whatever signs are 
given in a special case to the algebraic symbols in the 
formula. 

Convex Spherical Mirrors. When the mirror is convex, 
the point C, the centre, is behind the mirror, so that CA is 
negative. A mathematician would know without any further 

112 

proof that the formula - -f - = - must be true if the appro- 
p q r 

priate sign were given to r. As, however, all our readers 
are not necessarily mathematicians, we will give a formal 
proof of this case. 

As before, let C (Fig. 10) be the centre of curvature of 
the convex mirror AK, and let the incident ray PK be 



CONVEX SPHERICAL MIRRORS 23 

reflected at K in the direction KR, so that the angle of inci- 
dence = $'. Produce RK to Q, cutting the axis in Q. 
Then PKQ is a triangle, of which the exterior angle PKR 
is bisected by the line CK, that meets the base produced 
in C. 

. PK_PC_PA + AC_PA-CA_^-r 

QK " CQ ~ CA + AQ " CA - QA ~ r - q ( 

When, therefore, a thin centric pencil is under considera- 




FIQ. 10. 



tion, and PK and QK may be regarded as equal to PA and 
QA, or p and q, we have, as before 

P = P^L, which reduces to i + i = ? 

q r q p q r 

When the incident rays are parallel, i.e. when p = cc , 

1 121 

- = 0, consequently - = - = -?, so they are reflected as if 

P 2 T f 

they came from a point F behind the mirror, such that FA 

= JCA ; in other words, the principal focus is virtual and / 

1121 
is negative. We see, then, that the old formula -+- = - = - 

still holds good if we assign the proper negative value to 
r and /. 



24 GEOMETEICAL OPTICS 

Ex. An object is placed 8 ins. in front of a spherical 
mirror, and an image of it is formed 4*8 ins. behind the 
mirror. What is the radius of curvature of the mirror, and 
what is its focal length ? 

In this case p = 8 ins. and q = 4'8 ins., for the image 
is behind the mirror. So we get, on substituting these 
numerical values for the symbols, and paying due regard to 
the signs they bear 

1121 11 



/./= -12 andr= -24 

The negative signs show that both the radius and the 
focal distance are to be measured in the negative direction, 
i.e. both C and F are behind the mirror, that is the mirror is 
convex. 

This example will show the wealth of information that is 
contained in this simple formula. 

Geometrical Construction of the Image. Fig. 11 represents 
a concave mirror, the centre of which is at C, and an object, AB, 
on the principal axis of the mirror ; we are now going ibo show 
how the image ab can be constructed, assuming, of course, 
that only centric pencils contribute to its formation. As has 
already been pointed out, the previous formula is only true 
for thin centric or axial pencils, consequently in this case we 
shall draw a tangent plane HOH' to the vertex of the 
mirror, which will be a better representation of this centric 
portion than the whole curved surface of the mirror would 
be. In the figure, the line ACa may be taken to represent 
a thin centric pencil as it passes through the centre C, but it 
lies on a secondary, not on the principal, axis, and for this 
reason the term " centric " is less likely to be misunderstood 
than " axial." The- plane HOH' will in future be called the 
Principal Plane. We will give two methods by which the 
image a can be found of a point A that is not on the principal 
axis. 

(A) Point not on the Principal Axis. 



CONSTKUCTION OF THE IMAGE 



25 



(1) Draw ACa through the centre C, and draw AH to 

the principal plane parallel to the principal axis ; 
draw HFa through the focus to meet the line ACa 
in a. 

Then a, the point of intersection of HFa and ACa, 
marks the position of the image of the point A. 

(2) (Dotted lines.) Draw ACa through C, and draw 

AFH' through the focus to the principal plane; 
draw H'a parallel to the principal axis until it 
meets ACa in a. The point a is the image of the 
point A. 




FIG. 11. 

(1) The reason of this procedure will be apparent from the 
following considerations. We know that any ray parallel to 
the principal axis will be reflected through F, the principal 
focus ; consequently, if AH represent such a ray it must be 
reflected as HFa. Further, we know that any ray drawn 
through the centre C must be reflected back along the same 
course through C. Hence the point of intersection a must 
represent the image of A. It is true that the line AH does 
not represent any ray that is actually incident on the 
mirror, for the point of its incidence would be so eccentric 
that it would not be reflected through F. For this reason 
the eccentric part of the mirror (suggested by spaced lines) 



26 GEOMETRICAL OPTICS 

has been covered up. We are, however, justified in asserting 
that a small centric pencil from A will come to its conjugate 
focus at the point of intersection a of HF and AC. 

(2) We also know that light from any luminous point at 
F must be reflected back parallel to the principal axis; 
consequently the ray FH' must be reflected back as H'a, 
so that a, its point of intersection with AC, must be the 
image of A. 

A vertical plane at F perpendicular to the principal axis, 
such as FD, is called the Focal Plane, and has the following 
properties: Light from any point on this plane will after 
reflection travel in rays parallel to that axis on which the 
point lies. For instance, light from D will be reflected from 
the mirror as a beam of parallel rays in the direction H"5, 
which is parallel to DC, the axis on which D lies. Also all 
pencils of parallel rays that are but slightly inclined to the 
principal axis will after reflection intersect in some point on 
the focal plane. This property enables us to determine the 
position of the image of a point that lies on the principal 
axis. 

(B) Point on the Principal Axis. (Spaced and dotted 
lines.) 

Through B draw any ray BDH", cutting the focal plane 
in D; join DC and draw H"& parallel to DC, cutting the 
principal axis in b. Then the point I is the image of the 
point B. 

Precisely the same construction can be applied to the 
case of a convex mirror, which the reader is recommended to 
draw for himself. It will be noted that whenever the 
object is vertical, the image also will be vertical, so that 
in drawing the image of AB in practice, all one has to do 
is to find the position of a by either of the methods given, 
and then draw a vertical line from it to meet the principal 
axis in &. 

Size of the Image. From a consideration of Fig. 11 it 
is easy to find an expression for the height of the image (i) 
as compared with the height of the object (0). 



SIZE OF THE IMAGE 27 

(1) Noting that ba is equal to OH', we find by similar 
triangles that 

*___&_ OH; _FO_ FO / 

o ~ BA ~ BA ~ FB "" FO - BO ~/ -p 

(2) And, seeing that BA is equal to OH, it follows that 

i-*!L lOL F& - FQ ~ &Q -/ - 9 
o " BA " OH ~ FO " FO / 

The formulae which we have now proved, - 4- = 1, 
dealing with the position of the image, and these 



. 

o f-P f 

dealing with the size of the image, can hardly be overrated, 
since almost exactly the same formulae will be found to be 
true when we come to deal with refraction at spherical 
surfaces, while the method of construction of the image and 
the method of determining its size are merely a repetition 
of what we have just done. 

It should be noted that all erect images are virtual, while 
all inverted images are real. 

An example or two will show the value of these formulas. 

Ex. (1) A concave mirror has a radius of curvature 
of 10 ins. What is its focal length ? An object 4J ins. 
in height is placed 50 ins. :in front of the mirror. What 
is the height of the image, and where is it formed ? 

M 

Here r = 10, so/ = ~ or 5 ins. 


And since +=!, / = l _ / = ^"/ Q = fP 
p q q P P p-f 

(This is the best form to use in all cases, as a similar 
form holds good for refraction at spherical surfaces.) 



28 GEOMETKICAL OPTICS 

The position of the image is then 5| ins. in front of the 
mirror, since q is positive. 

Again, since - =- r ) - =--= - ^ = - = - - 
o f po 5 50 45 9 

The image is therefore inverted (as the sign is negative) 
and real, and it is the height of the object ; i.e. 



It is immaterial which formula for the size of the image 
we use. Let us try the other formula : 



So the height of the image is ^(4J), or ^ in. 

Of course, the formula about size refers also to width. 
Our second example is slightly more difficult. 

Ex. (2) An object 6 cm. in width is placed at a 
distance of 9 cm. from a reflecting surface. A virtual image 
2*4 mm. in width is formed of it. What is the radius of 
curvature of the reflecting surface ? 

Since - = ^ , if - ip = fo :.f(i - o) = ip 

and since the image is virtual, i is positive. 
Consequently 

2/* = 2i P - 2 x 2 ' 4 x 9Q - 4 ' 8 x 9Q 

' ' ~ i - o~ 2-4-60 -57'6 

48 60 



6'4- 8 

As the sign is negative, the surface must be convex. 

It may be noted that this is the basis of the method by 
which the radius of curvature of the cornea is determined in 
a living subject. A special apparatus is used to measure the 
size of the image reflected from the surface of the cornea, and 
from this measurement the curvature is calculated as in this 
example. 



GEAPHIC METHOD 



29 



Graphic Method for Spherical Mirrors. There is a very 
simple graphic method for finding the position and relative 
size of the image, which we will now give. The formula 

- -f 

= 1 is so similar in form to the well-known equation 



P 



x 



If 



to a line in terms of its intercepts - + | = 1, that it at once 

suggests the following device (Fig. 12). 

Draw two axes PH and HQ at right angles to each other, 
and consider H the origin. On PH mark off a distance F'H 
equal to the focal length / of the mirror, and draw the line 
FT also equal to/ at right angles to PH. 




FIG. 12. 

As with a concave mirror / is positive, F'H and F"F 
must be measured in the positive directions, i.e. either from 
left to right or from below upwards (standard notation). 

Suppose that an object is placed 20 ins. from a concave 
mirror of focus 4 ins., and we wish to determine the position, 
size, and nature of the image. Mark off the point P on PH 
so that PH = 20 ; join PF", cutting HQ in Q. 

Then QH is the distance of the image from the mirror ; 
as it is measured from below upwards QH is positive, and it 
is found to be 5 ins. in length, so the image is situated 5 ins. 
in front of the mirror. 

Again, *- =j ; but/ - p = F'H - PH = F'P 



i F'H , . , . ... . 4 

* - = ^F7r> > which in this case is ^ 
o r r lo 



As F'H and F'P are measured in opposite directions - 



30 GEOMETEICAL OPTICS 

must be negative, so the image is inverted and real, and its 
height is ^ of that of the object. 

With the convex mirror /is negative, let/= 4 ins., so 
FH and F"F are drawn as shown in Fig. 13, in the negative 
direction. Suppose the object to be 16 ins. from the mirror, 
PH is then made 16, and QH is found to be - 3'2 (negative 
because measured downwards). Consequently the image is 
situated 3*2 ins. behind the mirror. 

A A i F'H -4 1 

Andas _=_ = __=_ 

the image is virtual and erect, because - is positive (F'H and 




Fia. 13. 

FT being measured in the same direction), and its height is 
J that of the object. 

*Eccentric Pencils ; Spherical Aberration, In the preced- 
ing sections we have been considering the reflection of small 
centric pencils only ; when the incident light forms a wide 
cone the several reflected rays cross each other at different 
points, as is indicated in the diagram (Fig. 14). Paying 
attention first to the eccentric pencil that falls on the mirror 
at PQ, the reflected rays cross at FI and meet the axis in a 
line at F 2 nearer the mirror than I, the conjugate focus of S 
for centric pencils. The figure, of course, only represents a 
section of the whole mirror ; if we consider it to be rotated 
through a small angle about the axis OCS, the point FI will 
trace out a small arc (approximately a line) so that at 
FI and F2 two small lines will be formed : these are called 
the primary and the secondary focal lines. 

It will be noticed, also, that all the reflected rays that are 
incident on the mirror from S touch a certain caustic surface 



SPHEKICAL ABEKRATION 31 

which has a cusp at I. This caustic curve may be commonly 
observed on the surface of the fluid in a teacup, being formed 
by the light that is reflected from the inside of the cup. A 
caustic curve is frequently defined as the locus of the primary 
focal lines ; this only means that it is formed by the points 
of intersection of successive reflected rays. 

The term " spherical aberration " is applied to all the 
phenomena depicted in Fig. 14. If the reflecting surface had 
not been spherical, but had been part of an ellipsoid of 
revolution of which S and I were foci, all the light from S 
that fell on this elliptical surface would have been accurately 




FIG. 14. 



reflected to the conjugate focus I, and the surface would then 
be an " aplanatic " (airXavi'ig, not wandering) surface for the 
source S. Similarly, a paraboloid would be aplanatic for 
parallel rays. 

The " longitudinal aberration " (IF 2 ) of a thin eccentric 
pencil is the distance between the point where it cuts the axis 
and the focus for a thin centric pencil. 

Fig. 15 gives another view of the reflected pencil which 
is shown in section as the shaded area in Fig. 14. The 
incident cone is supposed to be pyramidal, so that the reflect- 
ing portion of the mirror concerned is of the rectangular 
shape PQKKi. The reflected pencil is of a very curious shape, 
the upper and lower rays (e.g. QFi and PF X ) of the two sides 



32 



GEOMETEICAL OPTICS 



intersect in the primary focal line at FI, while the rays QF 2 
and EF a from the upper edge intersect at F 2 the bottom of 
the secondary focal line, and PF 2 and EiF 2 from the lower edge 
at the top of the secondary focal line. Pencils like this 
which do not converge to a point but to two focal lines are 
called "astigmatic pencils" (a priv., crrfyjua, a point). As 




FIG. 15. 

we shall find that precisely the same condition arises with 
eccentric pencils when refracted, it is well to spend some 
little time in forming a clear conception of the solid figure 
represented between the two focal lines at FI and F 2 . In 
crystallography this sort of double wedge is called a sphenoid. 
Near F 2 its section would be oblong, its height much greater 
than its width; at D the section would be square, while 
nearer FI it would again be oblong, but with its width greater 
than its height. Clearly if the incident pencil were circular 



REFLECTION AT SPHERICAL SURFACES 33 

in section, the outline at D would be a circle. D is then the 
position of what is called " the circle of least confusion," and it 
is the best representative of the focus of an astigmatic pencil. 
We will conclude this chapter by giving the formulae 
without proof by which the position of the primary and 
secondary focal lines can be determined of an incident thin 
oblique or eccentric pencil on a spherical surface (Appendix, 
p. 109). Let SP be denoted by u, FI? by v l} and F 2 P by v 2 , 
and the angle of incidence by <. 

Then Ul.-JL and l - + L = 



u V rcos< u v r 



QUESTIONS. 

(1) An object 9 cm. in height is placed 10 cm. in front of a concave 
mirror of focal length 25 cm. What is the height and character of the 
image, and where is it formed ? 

(2) The radius of a concave mirror is 16 ins. What is the distance 
of the image from the mirror when the object is placed at a distance of 
12 ins., and when placed at a distance of 4 ins. ? 

(3) A virtual image one-fourth the height of the object is formed 
by a mirror. If the distance of the object be 9 ins. what is the radius 
of the mirror ? 

(4) An inverted image of a candle is thrown on a screen at a 
distance of 6 feet from a mirror of focal length 6 ins. Where is the 
candle placed, and what is the relative size of the image ? 



CHAPTEE IV 

REFRACTION AT PLANE SURFACES 

So far we have been considering the path taken by light as it 
travels through a single homogeneous medium ; we must now 
find out what happens when light passes through more than 
one medium. For the sake of simplicity, we shall limit our- 
selves to the consideration of homogeneous light, i.e. light of 
the same wave frequency or colour, traversing different homo- 
geneous media. When light passes from one medium into 
another, e.g. from air into glass, its course is altered, and the 
light is said to be refracted. The subject was experimentally 
investigated in 1621 by Willebroard Snell, who found that 
there are two laws governing refraction : 

LAW 1. The refracted ray lies in the plane of incidence. 

LAW 2. The sines of the angles of incidence and refrac- 
tion are in a constant ratio for the same two 
media. 

The first law requires no explanation after what we 
have already said about the similar law of reflection 

( P . ii). 

Students of physics will know that light in vacuo travels 
at a rate of more than 186,000 miles per second, and that in 
dense media its velocity is less. Indeed, in Optics the ex- 
pression dense medium merely means a medium in which 
light travels with a relatively slow velocity. For instance, 
water is dense as compared with air, but rare as compared 
with glass , the speed of light in water is about f of its speed 
in air, but about |- of its speed in glass. Now, it has been 
mathematically proved and experimentally demonstrated 
that the constant ratio in Snell's law is identical with this 

34 



KEFKACTION AT PLANE SUKFACES 



35 



velocity ratio. If we denote the angle of incidence by <j> and 
that of refraction by #', when light passes from water into 



air its course is so altered that 



sin 



, = , i.e. about f . 
v 



When light passes from air into glass, 









| approx- 



mately, for the speed of light in air is about f times its speed 
in glass. 

Let BB' (Fig. 16); represent the vertical plane that 
forms the boundary of a dense 
medium, glass for instance, and 
let SC be an incident ray, CQ the 
corresponding refracted ray, and 
let NCM denote the normal at 
the point of incidence C. The 
original direction of the incident 
light is indicated by the dotted 
line OS', so MCS' is the angle 
of incidence (ft, and MCQ is the 
angle of refraction ^'. Note that 
the angles of incidence and refrac- 
tion are to be measured from the 




FIG. 16. 



same normal, in this case CM. In the figure 



S'L 



and 



sinf ~QM' 
they are both positive, as of course they should be, for the 

ratio - ^7 is the ratio of the speed of light in the first 

medium to its speed in the second medium. As light is 
reversible, we see that an incident ray in the direction QC in 
the dense medium will be refracted at the surface BB' of the 
rare medium as CS. Indeed, it follows from Law (2) that 
all light on entering a dense medium is refracted towards the 
normal, and on entering a rare medium away from the 
normal. 

The ratio of the velocity of light in a vacuum to its 
velocity in a certain medium is termed the absolute index of 
refraction of that medium, and it is denoted by the symbol 
fi. Thus, if the absolute index of water be f and that of a 



36 



GEOMETRICAL OPTICS 



certain kind of glass be f , it is easy to determine the relative 

y w 
index of water and glass, or . 



339 
= X -T = x or 



sm 



4 8 



Snell's law may be then stated as T* y , = 2 , where the 

sin 0' /n' 

subscripts denote the first and second media ; when the first 
medium is air, -- 2. ^ 2 . The refractive index of air 

differs so slightly from unity (being less than 1*0003) that 
we shall commonly regard it as unity and denote it by JUQ. 

Ee turning again to Fig. 16, the angle S'CQ is called the 
deviation (D) of the light in this medium when the angle of 
incidence or is MCS'. As SCS' is the original course of the 
light, the deviation S'CQ or D is equal to - (MCS' - MCQ) 
or D = (0 $') . This is clear, for the new direction <j>' 
must be equal to (the old direction) + D. Annexed is a table 
in which the angles of refraction and deviation are given 
that correspond with certain angles of incidence at the sur- 
face of a kind of glass whose index of refraction ju is 1*52, 
and again at the surface of water when /* = T333. 



fA=l'52. 


p = 1-333. 


$ 


<t>' 


D 


I 


f 


D 


20 


13 0' 


- 7 0' 


20 


14 52' 


- 5 8' 


40 


25 1' 


- 14 59' 


40 


23 50' 


- 11 10' 


60 


34 44' 


- 25 16' 


60 


40 31' 


- 19 29' 


80 


40 23' 


- 39 37' 


80 


47 38' 


- 32 22' 


90 


41 8' 


- 48 52' 


90 


48 36' 


- 41 24' 



It will be noticed that when the angle of incidence in- 
creases uniformly, as in the first four rows, the angle of 
refraction increases slower and slower; consequently, the 
deviation ($' 0) increases faster and faster. It follows 
from what has been said before that if the light be considered 
as passing from the dense medium to the rare medium, we 



CRITICAL ANGLE 



37 



must consider the <' in the above table as the angle of inci- 
dence, and $ the angle of refraction. 

What will happen when the angle of incidence in water, 
for instance, is greater than 48 36'? When $ is nearly 
48 36' (Fig. 17) the refracted light just skims along the sur- 
face of the water. If be more than 48 36' the light will 
be unable to leave the water, and it will be totally reflected. 
The angle of incidence (48 36') at which this phenomenon 
of total reflection occurs is called the critical angle. 
It should be noted that when the path of light is from a 




Fio. 17. 

dense to a rare medium the relative index of refraction is 
less than unity, for d ju r = ; in the case where yu r = 1, 

d fjL r = -. The critical angle is easily found for any medium ; 
we have merely to give $' its maximum value of 90, and our 



formula tells us that 



-I; thus sin 



= df* r = -, " the critical angle 



= 48 36'. 



Total reflection is frequently taken advantage of in the 
construction of optical instruments, e.g. the camera lucida 
(Appendix, p./^f ) and prism binoculars. Similarly when 
the surface of water is viewed in a glass held above the 
level of the eye, the silvery brilliance of the surface is due 



38 



GEOMETRICAL OPTICS 



to total reflection. In aquariums the surface of the water is 
often seen to act as a brilliant mirror. 

We see, then, that part at any rate of the light in a rare 
medium can always enter an adjoining dense medium, but 
that when light in a dense medium is incident at any angle 
greater than the critical angle, none of it will leave the dense 
medium, as it will be totally reflected at the boundary of the 
rare medium. 

Image by Refraction at a Plane Surface. Let P (Fig. 18) 
be a source of light in a dense medium (//) that is separated 




FIG. 18. 

by a plane surface AB from a rare medium (JUQ), say air. 
Draw PAE normal to the surface. We will consider how the 
point P will appear to an eye at E. 

Since P is sending out light in all directions, we may 
regard PA and PN as two of them. The ray PA will 
undergo no alteration, for since PA is normal, both and 0' 
must be 0. Draw the normal HNK at the point of 
incidence N; then HNP = 0, and if HNQ be so drawn 

that sm JE^ = ?, HNQ must be <i>'. Under these cir- 
sin HNQ IUL 

cum stances produce QN to E, then the ray PN will be 
refracted as NR. Now, if the figure be rotated round EP as 



IMAGE BY KEFKACTION 39 

axis, PN will trace out the limits of the axial incident cone 
and QNK the limits of the emergent cone. Since HNP and 
HNQ are equal to the alternate angles QPN and AQN, 

o 1 _ sinHNP _ sin QPN _ sin QPN _ QN 
fi' ' r fj! ~ sin HNQ ~~ sin AQN " sin NQP PN 

As, however, we are only considering the narrow pencil that 
enters the pupil of the eye at E, the point N will be very 
close to A, and QN and PN may be regarded as equal to QA 
and PA ; consequently, to an eye at E the position of the 

"P A 

virtual image of P will be at Q when QA = ^?PA = . 

If, for example, a small object in water is viewed from a 
point immediately above it, its apparent depth will be f its 
real depth. 

As in the case of reflection at spherical mirrors, this 
determination of the situation of the virtual image is only 
true when small pencils that are nearly normal to the 
surface are considered. The formulae for oblique pencils are 
more complicated, as two focal lines are formed with a circle 
of least confusion between them (see Appendix, p. 111). 

Refraction through a Plate. Let us consider how an 
object P in air (JJLQ) will appear to an eye E in air when 
viewed through a plate of glass (JUL) of which the thickness t 
is AB (Fig. 19). The incident ray PN will be refracted as 
NM on entering the glass, and on again entering the air at 
M its course will assume the direction MR. As we are only 
considering what will be the position of the virtual image Q 
to an eye at E, we can find it very easily by the method of 
the last section. 

At the surface AN the point P will form a virtual image 

i 

at Q' such that Q'A = PA. But now Q' may be regarded 
l*o 

as the object for the second refraction at BM, and 
Q'B = Q'A -f t, consequently we have for the distance QB 
from the distal surface 



QB = Q'B = (Q'A + = PA + 



40 GEOMETRICAL OPTICS 

and since QA = QB - t, 



When /MO =1 and // is denoted by ju, QA = PA - 



\ 



and QB = PA 4- -. Thus if fi = |, QA = PA - 1 or the 

fj. A 6 

image is formed nearer the plate by ^ of its thickness. 




FIG. 19. 

It will be noticed that the emergent ray MR is parallel 
to the incident ray PN ; this is invariably true however 
many media there may be, provided that they are bounded 
by parallel planes, and provided that the final medium has 
the same refractive index as the initial medium. 

Succession of Plates. It is easy to see that, if any number 
of media bounded by parallel planes are in succession, light 
on emerging will pursue a course parallel to its original path 
if the initial and final media have the same refractive 
index pi. Let jui, ju 2 , ju 3 . . . denote the indices of the first, 
second, third . . . medium respectively, and let 0i, fa, 3 . . . 
denote the angles of incidence in these media ; then 



sn 



sn 
sin 



SUCCESSION OF PLATES 



41 



for, since the media are bounded by parallel planes, the angle 
of refraction at one surface is equal to the angle of incidence 
at the next. Thus, if there are four media interposed (Fig. 20), 
and the angle of emergence into the final medium be denoted 



.*. sin = sin b 
or, in other words, the final and initial rays are parallel. In 





FIG. 20. 

the diagram the third medium is represented as less dense 
than either of those adjoining it. Consequently, fa is greater 
than ^>3 and ^g. 

Prisms. Any refracting medium bounded by two plane 
surfaces which are inclined at an angle to each other is 
called a prism. The inclination of the faces BA and CA 
(Fig. 21) is called the refracting angle or the apical angle of 
the prism, and is usually denoted by A. The median vertical 
plane that bisects the apical angle is called the principal 
plane, while the plane CB at right angles to the principal 
plane is called the base of the prism. We proceed to 



42 GEOMETEICAL OPTICS 

demonstrate certain properties which are common to all 
prisms. 

I. When light passes through a prism which is denser 
than the surrounding medium, it is always deviated towards 
the base. 

If the face BM in Fig. 19 were rotated clockwise through 
a small angle about B, the plate would form a prism with its 
edge upwards and its base downwards ; the angles of incidence 
and of refraction at BM will therefore diminish, and conse- 
quently any incident ray NM will on emergence be deviated 
away from ME towards the base. If the rotation of BM be 
continued, the angle of incidence at M, after passing through 
the value of 0, will change its sign and become positive, 
when a still more marked deviation of the emergent ray will 
occur. Were the face BM rotated counter-clockwise, the 
base of the prism then formed would be upwards, and at the 
same time the negative angle of incidence would increase, 
and hence also the angle of refraction. The deviation of 
the emergent ray would in such a case be upwards, towards 
the base. 

If the prism be less dense than the surrounding medium 
the deviation is towards the edge of the prism. 

II. As the angle of a prism increases the deviation also 
increases. 

This follows immediately from the proof given of I. 

III. The apical angle is equal to the difference between the 
angles of refraction and incidence within the prism. 

When the apical angle A is measured in the same 
direction as the angle of deviation D, 

A = <// - $'. 

In all the books on Optics A is said to be equal to </>' + ^', and in 
order to get this result a new and special convention is adopted for the 
signs attributed to <>' and \J/, which is never elsewhere employed. Our 
convention about clockwise and counter-clockwise rotation when 
consistently used, will be found to give results that always hold good. 

Let SIET denote a ray of light passing through the prism 
BAG (Fig. 21). Draw the normals at I and E, and let the 



PRISMS 



43 



angles of incidence and refraction at I be and <}>', and the 
angles of incidence and emergence at R be $' and ^. Note 
that in the figure <j> and $' are measured clockwise and are 
therefore negative, while $' and $ are measured counter- 
clockwise and are consequently positive. 

In the triangle AIR the angle IRA = 90 - <//, and the 
angle AIR = 90 + $', for $' is measured clockwise. Now, 



FIG. 21. 



B 



since the sum of the interior angles of the triangle AIR is 
180, 

RAI + 90 - f + 90 + 0' = 180 

.'. A or RAI = $' - '. 



and 



carry, 



This is universally true whatever signs 
whether they be different or both the same. 

IV. The total deviation is equal to the difference between 
the angles of emergence and incidence less the apical angle of the 
prism. D = ty A. 

If the rays SI and TR be produced to intersect at D, the 
exterior angle D of the triangle DRI is equal to the two 
interior and opposite angles IRD and DIR. 



44 GEOMETRICAL OPTICS 

But IRD = i/. - V, and DIE or -BID = -(<j> - 0') 
.*. D = t - - (i// - 0') = ^ - < - A. 

It is clear that the total deviation D is the deviation of 
the first surface, (0 0'), added to the deviation of the 
second surface, <A $'. 

As SIRT is the path of the light, the angle D is positive 
(counter-clockwise), and A is measured in the positive direc- 
tion also (RAI). If the light were travelling in the direction 
TRIS the deviation D would be negative, and then A (or 
IAR) would be measured in the clockwise direction. As 
i and ^ would then be interchanged, it would be found in 
that case also that 

D = ^ - - A. 

V. When a ray passes symmetrically through a prism, the 
deviation is a minimum. 

A ray passes symmetrically through a prism when the 
angle of incidence is numerically equal to the angle of 
emergence irrespective of sign, i.e. when < = - $. 

If <f> increases, 0' increases also ; at the same time, how- 
ever, \j/' diminishes, and consequently ^ also. But as the 
deviation (0 - 0') increases faster than the deviation 
^ \j/' diminishes (p. 36), the total deviation must increase. 
If we consider the path of light reversed, it appears that 
when the angle of incidence is diminished the total deviation 
increases. Hence the symmetrical position is the position of 
minimum deviation. 

When the prism is in the position of minimum deviation, 
^ = - and $' = - <(>' ; 

.-. D = 2A - A, and A = 2^' 

sin sin ^ sin i(A -f- D) 

^ ij QJ r ' L i. 

sin $' sin ^' " sin ^ A 

This is the method used for determining the relative 
refractive index of any transparent substance. The angles 
D and A can be measured with great accuracy by means of a 



PRISMS 45 

spectrometer, and from these data the value of ju can be found 
by this formula. 

The prisms that are prescribed for spectacles are usually 
very weak, i.e. the angle A is rarely more than 5, and for 
these a simple approximate formula can be given for the 
amount of deviation they induce. When $ and ^' are both 
very small, sin $ and sin // may be replaced by ^ and if/, 
and the formula for the position of minimum deviation 
D = 2^ - A may be replaced by 2/j.\j/' A or (/* 1)A, since 
A = 2\j/'. This is a most useful formula if only used for 
weak prisms when placed in the " symmetrical " position ; 
thus, as the value of jj. for glass is little more than 1*5, we 
find that the deviation produced by such a weak prism is 
about half its apical angle. 

Image by Refraction through a Prism. We have so far 
been only considering the course of light refracted by a 
prism ; when we try to form some idea of the appearance of 
an object when viewed through a prism, we meet with several 
difficulties that will only be briefly mentioned here. 

We have seen that the deviation is not proportional to 
the angle of incidence, and as an object of appreciable size 
presents many points from each of which light is falling on 
the prism at a different angle, it will be seen that the matter 
is a very complicated one. 

In Fig. 22 an object Pp is supposed to be viewed by an 
eye in the neighbourhood of ET, the narrow pencil from P 
that enters the eye being represented in the position of 
minimum deviation. Clearly a virtual image of P will be 
formed at Q, where the prolongations of the refracted rays 
intersect. It is obvious that the image will be displaced 
towards the edge of the prism, and that it will seem nearer 
by one- third the thickness of the prism traversed (p. 40). 
Now, if we consider light emerging from p, it is clear that if 
it fell on the prism in the position of minimum deviation it 
would form a virtual image in the neighbourhood of q. But 
an eye at E could not receive this pencil ; the only light that 
would enter it must have had some such initial direction as 
px, and the final image of p would be indistinct, for the 



46 GEOMETRICAL OPTICS 

lowest ray of the pencil entering the eye from p will have 
undergone a smaller deviation than the uppermost one, 
and consequently a confused image of p will 
about q'. 




FIG. 22. 



It will be convenient to give a summary of the appear- 
ances of a square object when viewed through a prism in 
different positions. 

(1) When the plane of the prism is parallel to the plane 
of the object, the edge of the prism being upwards. 

The image is raised above the level of the object, the 
sides being more raised than the mid-part, so that the upper 
and lower edges appear concave upwards. 

(2) When the prism is rotated about a horizontal axis 
parallel to its edge. 

The image rises, and the height of the image is either 
diminished or increased according as the edge of the prism is 
turned towards or away from the observer. 

(3) When the prism is rotated about a vertical axis, its 
right side being turned away from the observer. 

The right margin of the image is raised above its left 
margin, so that the right superior and the left inferior angles 



ROTATION OF PRISMS 



47 




FIG. 23. 



are rendered more acute. If the rotation be in the opposite 
direction, opposite results occur. 

(4) When the prism is rotated about the sagittal line, i.e. 
about the visual line of the observer. 

The image rotates in the same direction, for, being always 
displaced towards the edge, it follows it in its rotation. Let 
a point of light be supposed to 
fall on the centre of the cir- 
cular screen depicted in Fig. 
23. When a prism that causes 
a deviation R is interposed 
with its edge to the right, the 
point of light is deflected to 
the periphery of the circle on 
the right. If the prism be ro- 
tated through an angle /o, the 
spot of light will move along 
the arc through the angle p. 
Consequently, the vertical displacement V will be R sin p, 
and the horizontal displacement H will be R cos p. A 
prism is frequently ordered by oculists in this way to correct 
a vertical and a horizontal deviation of the eye simulta- 
neously. Clearly, if V and H are known the required prism 
can be found, for R = \/V 2 + H 2 . Practically, it is found 
what horizontal and vertical angular deviations are required 
by the patient. Say that they are 9 and 0, then tan 8 = H 
and tan $ = V, and a prism of deviation D is ordered such 
that tan D = R, and set at such an angle p that 

V tan 
tan p = = - 
H tan 

In ophthalmic practice the prisms are so weak that we 
may replace the tangents of 0, 0, and D by the angles (of 
minimum deviation) themselves. 

The reader is urged to verify the statements made in this 
section by experiments with a prism, but a full explanation 
of his observations can only be obtained in more advanced 
treatises. 



48 GEOMETRICAL OPTICS 



QUESTIONS. 

(1) Show why a stick that is partly in and partly out of water 
appears bent at the surface of the water when viewed obliquely. 

(2) It is found that when a plate of glass 7*7 mm. thick is placed 
over a microscopic object the microscope must be raised 2 - 7 mm. to 
bring the object into focus again. What is the refractive index of the 
glass? 

(3) If the apical angle of a prism be 60, and the minimum devia- 
tion for a certain kind of light 30, what is the refractive index of the 
material of the prism for this light ? 

(4) A prism of small apical angle (2), with refractive index T5, is 
placed in water of refractive index ^. Show that its deviation is only 
about one-fourth of what it is in air. 

(5) When viewing a distant object, each eye of a patient is found 
to deviate outwards 1 44' (nearly V^) while the right eye deviates 
above the level of the fixation line of the left eye 1. What two 
prisms would entirely relieve this defect ? 



CHAPTEE V 

REFRACTION AT A SPHERICAL SURFACE 

WHEN dealing with reflection at a spherical surface (Fig. 9, 
p. 19), we considered the object P to be on the positive side 
of the mirror. It would clearly make no difference to the 
distance of the image from the mirror if the object were to 
the right instead of the left of the mirror, except that the 
image Q would then be situated to the right also. Since 
optical formulae are universally true whatever values are 
given to p, q, and r, when using the formulas it will generally 
be found convenient to regard -the direction of the incident 
light as the positive direction. Hence PA or p may be 
regarded as positive whether measured from left to right or 
from right to left, and QA or q will be positive when both 
Q and P are on the same side of the surface of the medium, 
but q will be negative when Q and P are on opposite sides 
of the surface. 

When finding any general formula, to avoid error, it is the 
simplest plan to use the " Standard Notation." As will be seen 
in the following two sections, there is no difficulty in obtaining a 
correct general formula when the object lies to the right of the 
refracting medium. 

We shall in Fig. 24 (and sometimes in future) consider 
the object P to be situated to the right of the spherical arc 
AK to familiarize the reader with the fact that the direction 
in which light is travelling is quite immaterial when dealing 
with such questions as these. 

Concave Surface of Dense Medium. Let P be an object in 

49 E 



50 



GEOMETRICAL OPTICS 



a medium whose refractive index is JUQ distant PA or p from 
a concave spherical arc, AK, bounding a medium of index rf, 
and let CA or CK be the radius (r) of the arc ; then, if CKP 




FIG. 24. 



be the angle of incidence ft, and CKQ be the angle of 
refraction ft', 

ft __ sin ft _ sin CKP 
fiQ ~~ sin ft' ~~ sin CKQ 

Now, in the triangle PKC 

PC - sin CKP _ sin ft 
CK~sinKPC~sinKPC 



and in the triangle CKQ 



QC! 
CK 



sin CKQ 
sin KQC 



sm ft 
smKQO 



[Note that as PC, CK, and QC are all measured in the 
same direction, the angles CKP, CKQ, and KQ C must all be 
measured in the same direction, whether positive or negative. 
In this case PC and CK are negative, so their ratio is equal 
to the ratio of the sines of two positive angles.] 

On dividing the first expression by the second, we get 

PC __ sin ft sin KQC 
QC = sin^'sinrKPC" 1 
PA - CA p - r 

n ^ f\K* - - 

*" QA - OA 2 - r ' 



^ PK 

sin KPQ "~ /u ' QK 
PK 



REFRACTION AT A CONCAVE SURFACE 51 

This is universally true when p = PA, q = QA, and 
r = CA, and when /IQ is the refractive index of the medium 
in which P is placed. 

When a thin centric pencil is considered, K must be very 
close to A, and in such circumstances the distances PK and 
QK may be regarded as equal to PA and QA, i.e. to p and q. 
We then have for a thin centric pencil 

p - r _ fi p 
q - r~ fjto' q 
or fiopq - fnqr = p'pq - ppr 



On dividing by pqr we have 




or 



If now the figure be supposed to rotate round the axis 
PCA, the extreme ray PK will trace out the limits of the 
incident cone from P, and it is clear that all the constituent 
rays of this cone will after refraction proceed as if diverging 
from the point Q ; in other words, Q is the virtual image of 
P. Indeed, the image of a real object placed in any position 
before the concave surface of a dense medium is always 
virtual. 

When the first medium is air, /UQ = 1, and we get the 
ordinary formula of the books : 



q p r 

This, however, is neither its simplest nor its most easily 
remembered form, but before proceeding further we will 
show that this formula applies equally to convex surfaces. 
A mathematician would know, from the method of proof, that 
the result must be universally true however the signs of p, 
q, and r were changed, but we feel that this statement will 
not be convincing to all our readers. 



52 GEOMETEICAL OPTICS 

Convex Surface of Dense Medium. As before, let PA = p, 
QA = q, CA = r (Fig. 25), and consider the incident ray PK 
having NKP for its angle of incidence, or < ; if then NKQ be 
the angle of refraction, or 0', 

i sin sin NKP sin PKG 



sin 0' ~~ sin NKQ "~ sin QKC 



Now, in the triangle PKC, 



^ 4.u 4. - i mm Q sin 

and in the triangle QKC, = 



[A warning must here be given about signs : note that PC 
is measured in the opposite direction to CK, so the angle 




FIG. 25. 

PKC is measured in the opposite direction to KPC, one 
being clockwise and the other being counter-clockwise, and 
a similar precaution is necessary about the angles QKC and 
KQC. Want of attention to points like these may give an 
entirely erroneous formula.] 

On dividing the first expression by the second, we get 

sin 



- 
QC sm^'sinKPC /uo'sin QPK 

PA - CA p - r _ ju' PK 

QA-CA ^7- 



Consequently, in the limiting case when K is very near 
to A 



REFRACTION AT A CONVEX SUKFACE 53 

p-r ^p p 
q-r f*> q 



or 



The formula for a convex surface is therefore identical 
with that for a concave surface, but whereas the latter makes 
incident rays more divergent (unless p < r), a convex 
surface always makes them more convergent, and this 
depends on the fact that in the case of the concave surface 
CA or r is measured in the same direction as PA, and in the 
case of the convex surface in the reverse direction. 

It will be noted that in both Fig. 24 and in Fig. 25 the 
image Q is virtual, and hence P and Q are not conjugate in 
the sense defined on p. 20. It would be found, for instance, 
that if in Fig. 24 the object were placed at Q the image 
would be virtual and situated somewhere between Q and C. 
As stated before, P and Q are only interchangeable or con- 
jugate when the image Q is real. 

In Fig. 25 we expect that if PA were increased a little so 
as to render the incident cone less divergent, the refracted 
ray KE would be parallel to the axis PAG. Now, if KR 
were parallel to the axis, QA would be infinite, so let us put 
q = oo in our formula. 



Then L_^ = ^_ or / = - r 

co p r fjL /io 

We see, therefore, that when the source of light is put at 
a distance p the emergent rays are parallel. This position 
is called the First Principal Focus, and is denoted by F, 
while the distance FA is denoted by/' instead of p'. We 

have then/' = 

Now suppose that the incident rays are parallel, or that 
p is infinite, 

K- ' uo = t^-^- or </=-/- 



q oo r 



54 GEOMETEICAL OPTICS 

The point at which incident parallel rays come to a focus 
is called the Second Principal Focus and is represented by 
F" in the figure, while the distance F"A is commonly 
denoted by /" ; so, replacing q' by its technical symbol, we 



M - Mo 

It should be noted that /" has the same sign as r, 
and indeed F" is seen to be on the same side of the 
surface as C ; F', however, is on the opposite side, as is 
obvious from the expression for /' being preceded by a 
negative sign. 

Now, since the effect of a concave surface is to render 
incident rays more divergent (unless p < r) t it will be clear, 




FIG. 26. 

on considering refraction at such a surface, that in order that 
rays may emerge parallel they must have been initially con- 
verging towards the point F' (Fig. 26). Consequently the 
First Principal Focus F' in this case is virtual. Again, the 
Second Principal Focus F" is also virtual, as incident parallel 
rays (the spaced and dotted lines) after refraction diverge as 
if they were proceeding from F". It will be noticed that 
here also/" has the same sign as r, while/' has the opposite 
sign. 

Sinc e ^L ,=- 7 ^-, /"-r=-/' 

p. - ^ M - MO 



KEFKACTION AT A SPHERICAL SURFACE 55 

and we see, when the vertex of the spherical surface is 
denoted by A, that both in Fig. 25 and in Fig. 26 

F'A - CA = AF' or -F'A, or/" - r = -/'. 

Note that both F' and F" are real when the dense 
medium presents a convex surface. 
Again, it is clear that 

= - for --= 



/HO fi fio HQ ILL - JUG 

We can now express our formula in its simplest form, 
for on dividing 

V_W> = H'-VQ by P'"" weget 
q p r r 

f f" 
J - + ^ = 1 
P <1 

f f" 
This form, + = 1, is analogous to the expression we 

have already found for reflection at spherical surfaces ; then, 
indeed, there was no difference between the positions of the 
first and second principal foci, as they both coincided in one 

f f" 
point. It will be found that y - + = 1 is much the easiest 

P <l 

form to remember, especially as we shall find that exactly 
the same formula holds good with lenses. 

Ex. (1). The refractive condition of the human eye may 
be very closely approximated by a convex spherical segment 
of radius 5 mm. bounding a medium of index f . Where 
will its focus be for incident parallel rays, and what will be 
the position of its First Principal Focus ? 

Here the first medium is air, so /uo = 1 and // = J, 



/ = , - == -y f = 20 mm. 
and /' = -?/" = -|(-20) = 15 mm. 



56 GEOMETEICAL OPTICS 

So F' is situated 15 mm. in front of the cornea, and F" 20 
mm. behind it. 

If the focussing mechanism were not called into play, 
where would the image be formed by such an eye of an 
object 16'5 cm. distant ? 

/',/"_! ./"_*-/' Pf" 

-f- x . . ui q -j,i 

P q q p p-f 

Here p = 165 mm. 

165 x -20 -330 00 

- 



Presuming the curvature of the cornea and the other con- 
ditions of the eye to be normal, we should infer that if it 
could not see an object distinctly at a greater distance than 
16*5 cm., it must be 2 mm. longer than normal. We have 
just found that a human eye that can see very distant objects 
like stars distinctly must have its retina 20 mm. behind the 
cornea. 

Such problems are delightfully simple and easy, but mis- 
takes are frequently made when the object is in the dense 
medium. By attending to the rule that HQ is used to indi- 
cate the index of that medium in which the object is situ- 
ated, all difficulty will be avoided. We will give an example 
of the way such questions should be treated. 

Ex. (2). There is a speck within a spherical glass 
ball (r = 4 cm.) distant 2 cm. from one surface. Where 
will its image be formed as seen from either side when 
fi = 1-5 ? 

The first point to remember is that the object will form 
an image, whether real or virtual, irrespective of the presence 
or absence of an eye to see it. Consequently, neglect the 
position of the observer, except in so far as it determines 
which surface of the sphere is being considered, and use the 
" Standard Notation " for signs. 

In this case the object P is in the dense medium, so JU Q 
= 1'5, and the final medium is air, so p' = 1. Suppose the 
ball to be so placed that the surface nearest to P is to the 
left ; call this side A, and the distal side B. 



REFRACTION AT A SPHERICAL SURFACE 57 
A. Then PA or p = -2, and since TI = -4, 



and 




The image is virtual, within the glass, 1*6 cm. from A. 
B. Here p' or PB = 6, and r a = 4 ; 



so now/ 2 f or -TL- = 12, while / 2 " = -8 



The image is virtual, and is formed at the distance of the 
diameter from B, that is at the surface A. 

In this case, as the second medium is to be regarded as 
the refracting medium, it is obvious that at the surface B the 
air presents a concave surface, and the incident divergent 
cone from P is rendered less divergent, for QB is greater 
than PB. Indeed, in every case (except when p < r, as in 
case A) the effect of refraction at the concave surface of a 
rare medium is to increase the convergence or diminish the 
divergence. This can easily be remembered by considering 
the case of a biconvex lens : convergence occurs at the first 
surface, of course, because refraction occurs at the convex 
surface of a dense medium ; and this convergence is increased 
by the refraction at the second surface, which presents the 
concave surface of a rare medium (air). When the rare 
medium presents a convex surface, divergence results with- 
out any exception. 

Geometrical Construction of the Image. The construction 
is almost exactly the same as that we employed before, when 
dealing with a spherical mirror. The only difference is due 



58 



GEOMETEICAL OPTICS 



to the fact that now F' and F" no longer coincide, but are 
situated on opposite sides of the refracting surface. 

As before, we give two methods to find the image of a 
point not on the principal axis, and one method for a point 
that is on the principal axis. As we are assuming that only 
centric pencils contribute to the formation of the image, we 




FIG. 27. 

draw the principal plane HOH" tangential to the spheri- 
cal surface at its vertex (Fig. 27). AB represents the 
object, BOG the principal axis, F' and F" the first and 
second foci of the spherical refracting surface whose centre 
isC. 

(A) Point not on the Principal Axis. 

(1) Draw ACa through the centre C. Through A draw 

a line parallelto the principal axis, meeting the 
principal plane at H. Draw HF"& through F" to 
meet the line ACa in a. 

Then a, the point of intersection of HF" and AC 
produced, marks the position of the image of the 
point A. 

(2) (Dotted lines.) Draw ACa as before through the 

centre C; from A draw a line through the first 
focus F' to meet the principal plane at H'. Draw 



CONSTRUCTION AND SIZE OF IMAGE 59 

H'a parallel to the principal axis until it meets 
ACa in a. 

The point a is the image of the point A. 

It would appear, then, that the incident diverging cone 
HAH' becomes, after refraction, the converging cone HaH'. 
This is not strictly true, for the line AH does not represent 
any ray that is actually incident on the refracting surface ; 
but we are justified in asserting that a small centric pencil 
from A will come to its conjugate focus at the point of 
intersection of HF" and AC produced. Again, we know 
that all rays that pass through F', such as AF'H', must, 
after refraction, proceed in a direction (H'a) parallel to the 
axis. 

(B) Point on the Principal Axis. (Spaced and dotted 
lines.) 

Through B draw any line BDH" cutting the first focal 
plane in D and the principal plane at H" ; join DC and draw 
H"6 parallel to DC, cutting the principal axis in &. 

Then the point b is the image of the point B. 

The reason of this construction is obvious from the pro- 
perty of the focal planes described on p. 26. Light from 
any point on this plane will, after refraction, travel in rays 
parallel to that axis on which the point lies. Now, DC is, 
of course, the axis on which D lies, and consequently the 
incident light ray DH" must, after refraction, take the 
direction H'7> parallel to DC. 

Size of the Image. In just the same way as we found the 
height of the image formed by reflection at a mirror, we can 

find - in this case. 

(1) Noting that la is equal to OH', we find, by similar 
triangles, that 

i = fa_ = OH' _ FO _ F'O /' 

o ~~ BA "" ~BA ~~ FB "" F'O - BO "f^^p 

(2) And as BA is equal to OH, 

!=!?L JOL F"fr _F"0-50 
o BA ~ OH " FO ~ F'O 



60 GEOMETEICAL OPTICS 

Exactly the same construction is used when the refracting 
surface is concave, in which case F" with C would lie on the 
object side of the refracting surface, and F' would lie on the 
opposite side. It should be noted, also, that when the refract- 
ing medium is dense and presents a concave surface the 
image is erect and virtual, but when the surface is convex 
the image is inverted and real as long as p is greater than /' ; 
if p =f the refracted rays are parallel, and if p </' they 
are divergent. We will give one or two examples where JJLQ = 1 
(the first medium), and fi = f . 

Ex. (1). A refracting surface presents a convex surface of 
radius 4 cm., and at a distance of 8 cm. in front of it is 
placed an object 5 cm. high. Give the position, the character, 
and the height of the image. 

We must first find the foci 

=M-=--= =12 cm. 



and /" = - /' = -|(12) = - 16 cm. 



The image is therefore formed 32 cm. on the object side of 
the refracting surface, so it is virtual. 

I 12 



_ 
o f'-p 12-8"" 

or if we use the second formula 

i_f"-q_ -16 -32 
~o~~f~ -16 

The same result must be obtained in either case : the 
image is erect because - is positive, and three times higher 

than the object. Consequently, the height of the image is 
15 cm. 

This is a case in which p is less than/', and the refracted 



GRAPHIC METHOD 61 

rays diverge, though less than the original incident rays, so 
a virtual image is formed at a greater distance than that 
of the object. 

When speaking of the effect of concave refracting surfaces, 
we said that incident rays were always made more divergent 
unless p were less than r. As an example, we will now 
take a case where p is less than r. 

(2) An object 5 cm. high is placed 3 cm. in front of a 
concave refracting surface of radius 4 cm. Where will the 
image be formed, and what will be its height ? 

In this case, as the surface is concave r is positive ; so 




The image is therefore situated 3*2 cm. in front of the 
surface, or 2 mm. further off than the object ; in other words, 
the divergent cone from each point of the object is made 
rather less divergent, 



and 



So the image is virtual and erect, and as the object is 
5 cm. high, the height of the image is 4 cm. 

Graphic Method for Refraction at a Spherical Surface. 
The graphic method that we employed for spherical mirrors 
can also be very conveniently used in this case. As an example 
we give in Fig. 28 the method as applied to the normal human 
eye. The complex system of the eye with its cornea and lens 
will be shown on p. 102 to be almost exactly equivalent, from 
an optical standpoint, to a single refracting convex spherical 
surface of radius -5'25 mm., with /' or F'H equal to 
+ 15-54 mm. and/" or F"H equal to -2079 mm. 

On the horizontal straight line PH we mark off the 
point F' so that F'H = 15*54 mm., and from F' we raise a 
vertical line such that FT = 20*79 mm., for we consider 



62 



GEOMETKICAL OPTICS 



lines measured from left to right and from below upwards 
as positive, and those in the reverse directions as negative. 
Where will the image be formed by such a refracting system 
of an object 8 cm. distant ? 

Make PH = 8J cm., and join PF" and produce it to 
cut HQ in Q. Measure QH; it is found to be 25-56 mm. 
The image is therefore formed 25'56 mm. behind H, which 
denotes the cornea. Now, the " standard " eye that can see 



F" 




FIG. 28. 

very distant objects distinctly must have its retina at a 
distance of 20*79 mm. from its cornea ; suppose that a cer- 
tain person could not see objects further off than 82*5 mm. 
(this case), then his retina must be 25*56 mm. from his 
cornea, or his eye must be 4*77 mm. too long, if his eye 
were otherwise normal. 

The size of the retinal image is given by our previous 

f , i F'H 15-54 

formula, - = ^^ = ^757. , so that the image is inverted 



and real, and a little less than one-fourth of that of the 
object. 

Further, our diagram will tell us what glass would correct 
this eye for distance. Spectacles should always be worn in 
the first focal plane of the eye, i.e. about half an inch in front 
of the cornea, so that F' marks the position of the correcting 



GKAPHIC METHOD 63 

glass, and it must be of such strength that parallel rays from 
a distant object should form their image at P. This means 
that the second focal distance (/") of the correcting glass for 
distance must be equal to PF', which is found by measure- 
ment to be 67 mm. approximately. Now, a lens of which 

1 000 
/" = 67 mm. is one of power ^=- y or nearly 15D, as 

will be shown in the next chapter. Indeed, if preferred, this 
division sum may be done graphically in this way. Draw a 
line F'A 50 mm. long in any direction that makes an acute 
angle with F'P (Fig. 28); mark off S on PF', making SF' 
equal to 20 mm. Join AP and draw SD parallel to AP ; 
then F'D in millimetres gives the power of the correcting 
glass in dioptres (see p. 80). For of course 

F'D _ F'A . F'D _ - 50 _ -1000 - - 
SF' " PF" l ' e ' "20 = "67" ~67~ 

Note that both F'D and F'A are measured in the negative 
direction. 

If the eye were 3'23 mm. too short, where should the 
object be to form a distinct image on the retina ? 

Make QH -17'56 mm. (i.e. -2079 + 3-23). JoinF'Q, 
and produce to meet the base line in P (Fig. 29). If accu- 
rately drawn, P will lie 84'46 mm. to the right of H. This 
means that PH is negative ; thus, unless very great focus- 
ing power were used, the eye would not be able to see any 
real object distinctly, for light would have to converge as if 
towards a point 8446 mm. behind the eye in order to come 
to a focus on the retina. It is unnecessary to determine the 
size of the image, as we have seen that no real object could 
be seen. The correcting glass must have /" equal to PF', 
which by measurement we find to be - 100 mm. 

*-=.+ 

Or if we wish to do this division graphically, we draw F'A 
50 mm. long in any direction, making an acute angle with 



64 



GEOMETRICAL OPTICS 



FP. We then mark off S between P and F', making SF 
equal to 20 mm. Then by joining PA and drawing SD 
parallel to it, cutting FA in D, we measure FT). We find 




A 

FIG. 



that F'D = 10 mm., so his correcting glass is 4-10D. The 
graphic method is given in this case to show the generality 
of the method, though here there is no special advantage in 
it. As before, we have 



F'D F'A 



F'D 



50 



PF" l ' e ' -20~ - 



- J ~ 



-1000 
-100 



*Eccentric Pencils Focal Lines. In the preceding sec- 
tions we have been considering the refraction of thin centric 
pencils, of which the central ray traversed the centre of the 
refracting surface. We will now take the case of a thin 
eccentric pencil that does not pass through the centre. 

Let be a luminous point on the axis of the concave 
surface whose centre is at C (Fig. 30). The thin pencil 
POQ will be refracted in the direction RR', so that the 
refracted rays, if produced backwards, will meet at FI, and 
will cut the axis in a line in the neighbourhood of F 2 , so 
that the refracted pencil will be astigmatic. If the incident 
pencil were of a pyramidal shape, the prolongations of the 
refracted rays would form a figure something like Fig. 15, 
only in the case shown in Fig. 30 FiP is greater than F 2 P. 
A similar sphenoid will be formed between the secondary 



REFRACTION AT A SPHERICAL SURFACE 65 

and primary focal lines, but the lateral surfaces of the 
refracted beam will now intersect in the secondary focal line 
at F 2 before the final edge of the sphenoid is formed at F x , 
the primary focal line. If OP were greater than ^(FxP), 
F 2 P would be greater than FX? (see Appendix, p. 116); in 
such a case the refracted beam would take exactly the form 
represented in Fig. 15. 

The formulae for the lengths FiP or Vi, and for F 2 P or v 2 , 
when the source of light is at a distance OP or u t will be 
merely stated in this place 

/u cos 2 ft' _ cos 2 ft __/*__!_ sin (ft - ft') 

r sin ft' 




FIG. 30. 

The proof of these formulae will be found in the Appendix, 
pp. 114-116. 

Conversion of Formulae of Refraction into those of Reflec- 
tion. On p. 35 it was stated that -. , { or )= ^, i.e. the 

sin ft \ HQ} V 2 

ratio of the speed of light in the first medium to its speed in 
the second medium. Now, when reflection occurs at a sur- 
face, it is clear that the speed of the light is unchanged in 
absolute amount, for it is in the same medium, but its direc- 
tion is reversed ; in other words, the sign is changed, so that 

7 



66 



GEOMETKICAL OPTICS 



for reflection we must consider V 2 as equal to Vi. We 

' "XT' ~\T 

have therefore or ^ = ~ = 1, so that every formula 
/"o V 2 - V i 

applying to refraction must apply to the similar case when 

reflection is considered, if we simply replace the expression 

i 

for the relative index of refraction by 1. Indeed, this 

/*o 

is one of the best tests we have to determine whether a for- 
mula for refraction is correct or not. Eeplace // by - 1 and 
IHQ by 1, and see if the correct formula for reflection is given. 
We will give a few examples to illustrate the generality of 
this method of conversion. 



Refraction. 


Reflection. 


sin <f> /i' 
sin </>' ~JL^ 


sin _ 1 


.,=-,- 


sin ^' 


/*' /*0 "' ~ Mo 


-1 l_-2 ^1^1_2 


gj) r 


2 P r 


1> 2 * 


/* 1 sin (<f> $') 


- 1 1 sin 24 


_ 2sin0cos<J> 


V 2 u ~ r sin <f>' 


v t u rain 
11 2 cos 


^> "~ r sin ^ 


v z + u r 


/u cos 2 ^' cos 2 <f> sin (^ ^') 


COS 2 COS 2 (ft 


2 sin ^ cos 


1). % 81H rf> 


v l u 
or 1 + 1 


r sin ^ 
2 


v,u~ 


r cos ^ 



The reader need not therefore burden his memory with 
any of the special formulae for eccentric reflection, as the 
conversion of the formulae for eccentric refraction is so easy. 
As another aid to memory, I may add that any formula for a 
curved surface becomes true for a plane surface if we make 
r = oo , for this is equivalent to making the curvature 0. 
For instance, taking our last illustrations for eccentric 
pencils 

fi cos 2 V _ cos 2 ft __ fji _ 1 _ sin(<ft $') 

~ 



u 



v<i 



u 



r sn 



REFRACTION AT A SPHERICAL SURFACE 67 
becomes, when r is made infinite 

COS 2 ft _/f_ _ 1 _ Q 
U V% U 

COS 2 ft' 

or #1 = UM a- 1 - and t^ = M^ 
cos^ft 

These are the distances of the primary and secondary 
focal lines when a thin oblique pencil undergoes refraction 
at a plane surface, fj, denoting the relative index of refraction 
(Appendix, pp. 111-113). 

QUESTIONS. 

(1) Assuming that the human eye is a simple refracting system, of 
which the first focal distance is 15 mm. and the second focal distance 
20 mm., where would the image be formed, and what would be its 
height, if an object 6 mm. high were at a distance of 150 mm. ? 

(2) What curvature must be given to the refracting surface when 
/* = J in order that the previous object at a distance of 150 mm. may 
form a real image at a distance of 20 mm. ? What would be its 
height ? Compare the size of the retinal images in the axial myopia 
(the first case in which the eye is long enough to form a distinct 
image of the object) and the curvature myopia (the second case). 

(3) If the back of a glass sphere Q* = f ) be silvered, where will be 
the image that is formed by one reflection and one refraction of a 
speck that is halfway between the centre and the silvered side ? 



CHAPTER VI. 

LENSES. 

ANY refracting medium bounded by two curved surfaces 
which form arcs of spheres is known as a spherical lens; 
the axis of the lens is the line joining the centres of the 
spheres, and that part of the axis lying between the two 
surfaces gives the thickness of the lens. 

Thus in Fig. 31 the surface A is the arc of a sphere that 
has its centre at C b and the surface B is the arc of another 




FIG. 31. 

sphere that has its centre at C 2 : the axis of the lens is the 
line joining Ci and Cg, and the thickness of the lens is the 
length of the line AB. These are the only definitions we 
shall require for the present. It will be apparent that if one 
of the surfaces be plane, it may be regarded as a spherical 
surface of which the radius is infinite. 

Thin Lenses. Thin Axial Pencils. The conjugate focal 
distances of an axial pencil of a thin lens can be very 
easily determined from the formula that we have already 
found for a single spherical surface. It should be noted that 
the term " centric " is no longer equivalent to the term 
" axial," as an oblique pencil that passed through the centre 
of one spherical surface would not pass through the centre of 

68 



LENSES 69 

the other surface unless they coincided (see p. 84). We 
shall eventually determine the position of what is called the 
Optical Centre of a lens, and discuss its properties; at 
present we are only concerned with axial pencils. 

Let P (Fig. 31) be a luminous point on the axis of the 

lens, and let Qi be the image of P due to refraction at the 

r 

first surface A, and let be the relative refractive index 

Mi 

between the first medium and the second medium (e.g. air 
and glass). When p = PA, qi = QiA, and when r x is the 
radius of the first surface, 



qi p n 

The light from P, after traversing the first surface A, will be 
travelling in the direction towards Qi. On emerging at the 
second surface into the original medium (JUQ), refraction again 



takes place, and the relative index is now ^? . As the lens is 

considered to be of negligible thickness, QiB may be regarded 
as equal to QxA or q im 

Eegarding, then, Q! as the source for the second refraction 
at B, of which the radius is r 2 , and putting q for QB, the 
distance of the final image, we have 

i f 

JUQ JH fJ.Q fj. 

q qi r z 

On adding (a) t_^P = ^Jfo 

Mo fto , t 
we get = (ju 

This is the standard formula for a thin lens ; when the 
lens is in air (as is almost always the case) /^ = 1, and we 
can suppress the dash and write 



1 1 T /l 1\ 

=-!( ) . . . . (c) 

q p \n r 2 J 

To find the First Principal Focus (F r ), as before, we make 



70 GEOMETEICAL OPTICS 

the emergent rays parallel, or we make q infinite ; we then 
have - = 0, and writing / ' for this special value^ of p } we 
get 



To find the Second Principal Focus (F"), or the focus for 
incident parallel rays, we make p infinite ; so, substituting 
/" for this special value of q, we get 



/" 

We see that /' = -/", which is always the case when the 
initial and the final media are the same ; so, on substituting 
these values in (c), we find that 



111 /" /" 
= or ' *. = l 

q p f" q p 

and finally our old formula 

+^=1 (A) 

p q 

[It may be here noted that when thick lenses are con- 
sidered, if the thickness be denoted by t, the term 

must be added to the expressions for ry, and subtracted from 
that for ^7 . 



Thus 



and 






These expressions give the focal lengths correctly, and for 
some purposes this is all that is required, but they do not 



LfiNSES 71 

enable us to determine the position of F and F". We shall 
find later (p. 91) that when a thick lens is considered, there 
are two Principal Points towards which the two focal dis- 
tances must be measured respectively.] 

Let us suppose that the lens in Fig. 31 is of negligible 
thickness and has for its radii of curvature r x = 2 cm. and 
r a = 4 cm. What will be its focal distances when /u = 1*5 ? 

1/1 



/./' =f cm. and /" = f cm. 

Consequently, if a source of light were placed at 2| cm. 
distance from such a lens, the emergent rays would be 
parallel. If, however, the incident rays were parallel, as for 
instance from the sun, they would converge to a focus 2 cm. 
on the far side of it, as the negative sign shows that the 
Second Principal Focus is behind the lens. 

Ex. (1) If an object were placed 24 cm. in front of such 
a lens, where would the image be formed ? 

Here p = 24, and we wish to find the corresponding 
value of q. 

f+r =l ,r = P^L or q= *r 

p q q - p p-f 

' 24(-) 24 x 8 

80 2= 2T^| = -~6r~ = m - 

The image would therefore be real, and it would be formed 
3 cm. behind the lens. 

(2) An object is placed 24 cm. in front of a concave lens, 
of which TI = 2 cm. and r a = 4 cm. (n = 1*5). Where 
will its foci be situated, and where will the image of the 
object be formed ? 

This lens has the same curvature as that in (1), but in 
the reversed direction, so/' will be found to be f cm. and 



Hence q or = = 24 cm. 



72 GEOMETKICAL OPTICS 

The image in this case is virtual, and is formed 2*4 cm. 
in front of the lens. 

It should be noted that in all converging lenses /' is 
positive (i.e. F' is on the same side of the lens as the object 
P) and /" negative, and they both are real, whereas in all 
diverging lenses /' is negative (i.e. F' is on the side opposite 
to P) and / " positive, and they both are virtual. Moreover, 
all converging lenses are thickest in the middle, whereas all 
lenses which are thinnest in the middle are diverging in 
function. The reverse of these statements is not true, 
although it is frequently alleged to be so ; e.g. a lens can be 
constructed that is thickest in the middle and yet be diverg- 
ing in function. 

When a lens is converging in function, the image is real 
and inverted (Fig. 32) or virtual and erect (Fig. 33), accord- 
ing as the distance of the object is greater or less than the 
first focal distance of the lens. With diverging lenses the 
image of a real object is always virtual. 

In Figs. 32 and 33 the object is represented to the right 
of the lens, and therefore F' must be on the same (the 
incident) side, and as the lenses are of the same focal length 
(/' = 4 cm.) we make OF' in each case equal to 4 cm. In 
Fig. 32 the object AB is placed 14 cm. from the lens, so the 
image ab will be real and inverted, and it will be situated at 
a distance of 5'6 cm. on the other side of the lens, 

for 



. 
p~f 14-4 

In Fig. 33 AB is placed 3 cm. from the lens, and the 
image ab is therefore virtual and erect, and it is 12 cm. from 
the lens on the object side. 



P -/ 3-4 

Geometrical Construction and Size of the Image. The 
image is drawn ' by the same method that was employed in 
constructing the image formed by refraction at a single 
spherical surface (Fig. 27, p. 58). Three alterations are, 



CONSTRUCTION OF THE IMAGE 



73 



however, necessary : (1) the principal plane HOH' must be 
drawn through the centre of the thin lens instead of tangen- 
tially to its surface ; (2) OF" must be made equal to FO ; 
and (3) the point must be regarded as taking the place 
of C, which is no longer required (cf. Figs. 32 and 33). 




FIG. 32. 

When the lens is symmetrical, as in Fig. 32 and 33, 
O is the optical centre of the lens, and, clearly, any ray of 
light that traverses proceeds on its course without devia- 
tion, for any refraction that it may undergo on encountering 
the first surface of the lens will be reversed on emerging 
from the second surface. It is clear from Fig. 32 that the 
ray AOa cuts the lens at two points such that the tangents 
at the points of entry and emergence are parallel, so that 
the light traverses the lens as if it were a plate with parallel 
sides. 

(A) Point not on the Axis. The line AO is drawn and 
produced to a. (1) AH is drawn parallel to the axis, meet- 
ing the principal plane in H ; HF" is then drawn through F" 
meeting AOa in a (Fig. 32) ; or (2) F ; A is joined and pro- 
duced to meet the principal plane in H (Fig. 33), and from 
H the line Ha is drawn parallel to the axis, meeting OAa 
in a. 



74 



GEOMETRICAL OPTICS 



(B) Point on the Axis.From B a line BDH' (Fig. 32) is 
drawn at any acute angle cutting the first focal plane in D 
and the principal plane in H' ; DO is then drawn, and H7> is 
drawn parallel to DO. 

From Fig. 32 we see that 

F"Q - t>0 _f" ;- q 
F"0 /" 




FIG. 33. 



F'O 



and from Fig. 33 that 

i _ la _ OH _ F'O 

o ~~ BA ~ BA " F'B ~ F'O - BO ""/' - p 

Therefore when F'O or/' = 4, BO or^ = 14, and when BA 
is 8 inches in height, as in Fig. 32, 



f 4 

i or la = ^T- BA = T 

f-p 4-14 



8= -3-2 



But if BO or p = 3 and BA = 2, as in Fig. 33, 

^2 = 8 



f f" 

The formulae J - + = 1 

p q o j -p j 

thus shown to be universally true for refraction at a single 



SIZE OF THE IMAGE 75 

spherical surface, for lenses, and also for reflection, when it 
is remembered that in the case of a mirror F' and F" coin- 
cide in one and the same point F. 

Figs. 32 and 33 show, however, another expression for 

- that is always true in the case of lenses, for ^-r- = ^^ = - 

1>A JhJU p 

It is, however, rarely necessary to pay any attention to this 
relation, and it is much easier to remember the few formulae 
that are universal, which are given above. 

There is, however, one important point to which atten- 
tion may be particularly directed, as it affords the explana- 
tion of certain facts which are obvious enough if it be borne 
in mind, but puzzling otherwise. It is this, that the angle 
BOA subtended by the object at the centre of the lens is 
always equal to the angle bOa subtended by the image. The 
application of this fact may be illustrated by the following 
example : the diameter of the sun subtends a visual angle of 
31' : what is the diameter of its image as formed by a lens of 

1 m. focal length ? (tan 31' = 0'009.) 

In this case, of course, an inverted image is formed at 
F", and 

la or i = /" tan 31' = - 1000 X 0'009 mm. = - 9 mm. 

In the case of a mirror, it may be noted that the angle 
bOa is numerically equal to the angle BOA, as may be at 
once seen by drawing the lines aO and AO in Fig. 11. 
Then if bOa be denoted by a, BOA = a, for it is measured 
in the reverse direction to bOa, so that i = q tan a. When 
distant objects like the sun or the moon are observed with 
a reflecting telescope, the incident rays are parallel, and the 
size of the image formed by the mirror is given by the 
expression 

i = -/ tan a. 

Ex. (1). A camera of 6 inches focus shows a distinct 
image 1 inch high of an object when the ground-glass screen 
is 6*6 inches away from the lens. At what distance is the 
object, and what is its height ? 



76 GEOMETRICAL OPTICS 

Here an inverted image is formed 6*6 inches behind the 
lens, so 



:.p or **~*s = --g- g , c = 66 ins., i.e. 5 feet 6 ins. 
' 



q = -6'6 ins. 

**~*s = --g- g , c 
q j o'b + o 

A i /"- - 1 -6 + 6-6 1 

and o = V' ^T -=6- = ~io 

/. = 10 ins. 

Ex. (2). The same camera shows an image 1 inch in 
height of a man who is 6 feet high. How far off is he ? 

i f -16 

_ v m* _ __ _ 

o~f-p 72 ~G-p 

:.p - 6 = 6 x 72 ins. or p = 36J feet. 

Ex. (3). If in repairing a bicycle reflex lamp the plane 
mirror is placed at the focus of the convex lens, will it act 
in the desired way ? No. The light from a distant approach- 
ing motor that is incident upon the lens will converge to its 
focus. From this point on the mirror the reflected light will 
fall again on the lens, and will emerge as a parallel beam by 
its previous path. It will therefore only return to the 
source of light, and hence will give no warning to the driver 
of the car. 

It is for precisely the same reason that the pupil of the 
eye appears black. It is really red when incident light 
falls upon it ; but the observer necessarily puts his head in 
the path of the incident light in one direction, so that no 
light is returned in that direction. 

In order to see the back of the eye, an ophthalmoscope is 
used, which consists essentially of a small mirror with a 
central perforation, through which the surgeon looks at the 
eye of the patient. Light is reflected by this mirror into the 
patient's pupil to the red background of the eye, and return- 
ing by its previous path to the mirror is received in part by 
the surgeon's eye behind the aperture, so that the pupil 
appears of a bright reddish colour to him. If the retina of 
the patient be not situated at the focus of his eye, the light 



MAGNIFICATION 



77 



on its return from the retina will no longer emerge as a 
parallel beam, but either as a convergent or divergent pencil, 
according as the retina is behind or in front of the focus. 
In such cases, especially when the pupils are widely dilated, 
they will appear to glow with a ruddy light, even to the 
unaided eye. For a similar reason, the unsatisfactory reflex 
lamp may be made efficient by moving the plane mirror a 
little nearer the lens, so that it is just within its focal 
distance. 

It has been shown that in Fig. 33 the image ab is four 
times the height of the object AB, but this does not neces- 
sarily mean that the apparent size of this virtual image, to 
an eye that perceives it, is four times the apparent size 
of AB. 

Magnification. The apparent size of an object depends 
upon its distance from the eye that perceives it. If we 
regard the eye as a simple refracting medium formed of a 
single spherical surface of radius 5*25 mm., as on p. 61, 
the centre of this surface will represent the nodal point (K) 
of the eye 5 '25 mm. behind the cornea. Now, an object BA 



FIG. 34. 

(Fig. 34), at a distance KB from the nodal point, will sub- 
tend at K the angle BKA or 9. This is called the visual 
angle. The apparent size of BA is evidently determined by 

T> A 

tan or .- It is obvious that tan 6 could be increased 



indefinitely by diminishing KB indefinitely. In other words, 
the apparent size of an object could be indefinitely increased 
by bringing the eye indefinitely close to it. The eye, how- 



78 GEOMETRICAL OPTICS 

ever, is incapable of seeing an object distinctly which lies 
within a certain distance. The distance KB of this punctum 
proximum, as it is called, from the eye varies in different 
individuals, and increases with age, so that it is impossible 
to assign to it any definite value which shall be applicable to 
all cases. If I denote the least distance for the individual 
eye considered, then the greatest value that tan 6 can actu- 
ally have is by making KB equal to I, when tan 9 = -j- or y 

It is customary to assign an arbitrary value of 10 inches to I. 
If, now, a convex glass be placed as in Fig. 33, with the 
object BA within its first focal distance, and the nodal point 
of the observer's eye be at K, a virtual image la will be 
formed at a distance "Kb from the nodal point. Then if 0' is 

the visual angle subtended by la at K, tan 6' = T^T or ^. 
But we have already found that the maximum size of the 
object as seen by an unaided eye is given by tan 6 or j- 

The magnification M of a convex glass must therefore be the 
relation between tan 0' and tan 9. 

_ tan 0' _ i I _ l_ /l-J^JL/'i L^\ 
~ tan ~~ K6 ' o ~ Kl ' /" ~ Kl\ f) 

When a convex lens is used as a magnifying glass, the 
image is always virtual, and it therefore lies on the same 

side of the glass as F', so the fraction ~ is always positive 

f' ~ FO = OF"')' ^ * S C * ear k m ^ e a ^ ove ex P ress i n 
that M is increased by making KZ> as small as possible, and 
by increasing the value of q. But K& cannot be made 
smaller than I (say, 10 ins.), or the image would not be seen 
distinctly, and as "Kl = KO + 01, say, d -j- 05, there is a 
limit to the value of 01. The distance d of the lens from 
the nodal point of the eye cannot be much less than half an 
inch, so we must make 01 = 9 J inches in order to get the maxi- 
mum magnifying power out of a lens employed in this way. 



MAGNIFICATION 



In the case above, with a lens of 4 inches focus, with the 
object so placed as to form a virtual image 9J inches from 
the lens, the greatest amount of magnification will be 
obtained, viz. 



M = 



There is another way in which a convex lens may be 
used as a magnifying glass, viz. when the object is placed in 
the first focal plane of the lens (Fig. 35). HOH' is the 



FIG. 35. 

principal plane of the lens, which it is unnecessary to indi- 
cate, and the object BA is placed in the first focal plane, so 
that the incident cone of rays from B (H'BO) will, after 
refraction, emerge as a beam parallel to the axis, while the 
incident cone from A (OAH) will emerge as a beam parallel 
to AO. If, now, an emmetropic eye (or an eye adapted to 
see very distant objects) be situated behind the lens, an 
image of BA will be formed on its retina. It is clear that 
in this case the apparent size of BA will be independent of 
the position of the eye, for its size depends on tan BOA or 
tan 0". If, however, the eye be situated behind the point X, 
it is obvious that it could not see the whole of the object, as 
only rays from the central part of BA would enter the eye ; 
in fact, on increasing the distance of the eye from the lens 
the field of view would be diminished, but there would be no 
alteration of the magnification of that part that was seen. 



80 GEOMETEICAL OPTICS 

As BA is in the first focal plane, tan 0" = j, and so the 

tanfl" o I I 

magnification M in this case is - s = ~ f .- = JF . 

tan / o / 

Our lens of 4 inches focus, used in this way, would only 
give a magnification of j, or -^ = 2*5, but it would be less 

fatiguing to the eye. 

On comparing the magnifying power of a convex lens 
used in these two different ways, we see that in the first case 

it is 1 -f j n and in the second case it is j, or , ^. There- 
fore, if d be less than /' the first method gives the higher 
magnification, and vice versd. 

Graphic Method for Lenses. A method similar to that 
illustrated in Figs. 28 and 29 may be employed for finding 
the position and the relative size of the image formed by any 
lens. Fig. 28, when F"F' is made equal to F'H, will represent 
the construction necessary for a convex lens whose first focal 
distance (/') is indicated by F'H, whereas Fig. 29 if inverted 
would give the construction for a concave lens when F"F' is 
made equal to F'H, its focal distance. Such methods will, 
however, be seldom found to be of practical use, as we have 

vf " i f 

such simple formulae as a = * J ^ and - = -^ ready to 

P~f o f - p 

our hand for the solution of questions like these. 

Power of a Lens. Dioptres. In a previous section we 
used the expression " magnifying power," and we saw that 
itivaried inversely as/' ; we have seen, too, that strong lenses 
with strongly curved surfaces had short focal lengths. We 

are therefore quite ready to admit that -p is an adequate 

j 

measure of the power of a lens. The unit universally adopted 
is that of a lens of one metre focal length, which is called 
a dioptre, and is denoted by the symbol D. We know that a 
convex lens of 25 cm. focal length is four times stronger 
than one of 100 cm. focal length. This is very simply 
expressed by calling the former lens +4D and the latter 



LENSES 81 

+ 1D. Note that it is the first focal length, /', that is 
considered and that determines the sign of the lens. We 
give an example or two, so as to make this nomenclature 
quite clear. 

What is the power of the following lenses in dioptres ? 
two concave lenses of 10 cm. and 20 cm. focal length 
respectively, and one convex lens of 80 cm. and another 
convex lens of 22*5 inches focal length. 

The first focal distance of a concave lens is negative, 

so in the first case /' = -10 cm., or ^ metre; y, is 
then 10D ; in the second case/' = 20 cm. = - 5 metre, 
so , is -5D. 

The first focal distance of a convex lens is positive, so 
if /' = 80 cm., the power of the lens in dioptres is J^f, or 
+ 1-25D. 

In every case the dioptric power is given by -j, in metres, 

100 
or j-r- in centimetres. When we have /' given in inches, 

we must convert it into metres. As 39*37... inches are 

22 '5 

equivalent to one metre, 22 '5 inches are equivalent to QQ . Qf - 

t>y 'of 

1 39*37 

metres, so^,in metres is = 1*75. The dioptric power 

J Z&'d 

of this last glass is then +175D. 

Thin Lenses in Juxtaposition. A succession of thin lenses, 
such as that shown in Fig. 36, has practically the same effect 
as that of one lens whose power is represented by the sum of 
the dioptric strengths of the components of the system. Let 
us suppose that the first lens (a meniscus) is -I- ID, the second 
lens H-10D, and the third lens -9*5D; the sum of the 
dioptric strengths is +1 -f 10 9*5, or +1'5D. Conse- 
quently, the system will be practically equivalent to one 
lens of 4-l'5D. This illustrates the extreme simplicity 
that results from denoting lenses by their power instead of 
by their focal length. If the thickness of the lenses, or of 

O 



82 



GEOMETKICAL OPTICS 



their combination, had to be taken into account, a correction 
would have to be added, which will be explained when 
dealing with cardinal points. 




FIG. 36. 

When a space separates the several lenses of the system, 
the rule just stated does not apply ; for such a case the. 
reader is referred to p. 91. 

Optical Centre. We have frequently made use of the 
term " the centre of a lens," and the reader most probably 
,o thinks that it is equivalent to the mid- 

\ point of the thickness of the lens. It is 

so in symmetrical biconvex or biconcave 
lenses, but it may be outside the lens 
altogether, as in Fig. 37. The optical 
centre of a lens may be defined as that 
point in which the line joining the ex- 
tremities of any parallel radii of the two 
bounding surfaces cuts the axis. 

In Fig. 37, BACiC 2 is the axis, and 
and C 2 J 2 are two parallel radii. 




FIG. 37. 



The line joining J 2 and J b if produced, cuts the axis in 0, 
which is then by the definition the optical centre. 

The point is a fixed point, the position of which 
depends only on the lengths of the radii and the thickness 
t of the lens ; for by similar triangles 



Also 



C 2 O 

ri - Cl Tl ~ A - A 
^ - 0,0 = r a -OB ~ OB 

OB - OA AB = r 2 - 
"OB OB r a 



OPTICAL CENTEE 83 

So OB= ^- and OA = -^~ (a) 

r 2 - n r 2 - T*! 

Consequently, in biconvex and biconcave lenses, when either 
TI or r 2 is negative, r 2 - n > r^ so AB > OB ; in other 
words, is within the lens. The optical centre has the 
following important property : any incident ray, such as PJ 1} 
passing through the lens so that its direction while within 
the lens passes (either actually or virtually) through the 
centre will on emerging from the lens, have a direction 
J 2 Q parallel to its direction PJ X when incident ; and, con- 
versely, if any emergent ray be parallel to its corresponding 
ray, it must, while within the lens, have assumed a direction 
that would pass through the optical centre. This property 
of the optical centre follows at once from the fact that the 
tangents at the two points where refraction takes place are 
parallel, and therefore the effect on this ray is the same as that 
due to refraction through a plate (p. 40), i.e. that the angle of 
emergence is always equal to the angle of incidence 0. Pencils 
that pass either actually or virtually through the optical 
centre are called centric pencils. 

It is an easy matter to find the optical centre of any 
lens from the expression (a), if we are given the thickness 
t, (AB), and the radii of the two surfaces. It is quite 
immaterial which way the light is supposed to be travelling, 
whether from P to Q or from Q to P (Fig. 38). In the case 
of a biconvex or a biconcave lens, the ray while within the 
lens actually (not virtually) does pass through 0. 

Let us consider the left surface as A, the first surface, 
then if T-! = 3, r 2 = 4, and t = 2, by (a) we know that 



and - 

r a - 7*1 - 4 - 3 7 

and we notice that is situated within the lens and nearer 
the most strongly curved surface. Now, an incident ray PK' 
that emerges in a parallel direction, as K"Q, will not follow 
the course indicated in Fig. 38 within the lens, for it will be 



84 GEOMETRICAL OPTICS 

bent at the first surface, then pursue a straight course 
through 0, and then be again bent on emergence in the 
direction K"Q. It is, however, very convenient for purposes 
of calculation to find two points, K' and K", on the axis 
which will save us the trouble of calculating the angle of 
obliquity of the path within the lens. This can clearly be 
done by continuing the incident and emergent rays until 
they intersect the axis in K' and K", or the nodal points. 
This is a simple geometrical method, but it does not readily 
lead to an analytical expression, and the accuracy of the 
result obtained by a geometrical method depends, of course, 
upon the accuracy and size of the drawing. There are, how- 
ever, two analytical methods for locating the nodal points 




FIG. 38. 



that we shall presently give, but we shall first deal with a 
case in which the nodal points coincide at the optical centre. 

* Refraction of a Sphere. Coddington Lens. The sphere 
may be considered as a kind of double convex lens, and 
there are certain advantages attending its use, which we 
shall investigate. 

Note, first, that all pencils (even those that are oblique) 
which traverse the centre of the sphere pass normally into 
it, so that in this case all centric pencils are also axial. 

It may be readily seen, from (a), p. 83, that the geo- 
metrical centre of the sphere is also the optical centre of 
the lens ; so it will be convenient to find an expression for 
the focal distances when considered as measured towards the 
centre. 



SPHERE 



85 



Let (Fig. 39) denote the centre of the sphere, and let 
P denote an object in front of it, Qi its image due to 
refraction at the surface A, and Q the final image due to the 
refraction of Qi at the second surface, B. 

Let P denote PO = PA - OA = p - r lt 
and q r QiO = Q X A - OA = ^ - r b 
and let Q QO = QB - OB = q - r 2 . 
By the formula of p. 53, 



or 




Dividing by p' 



Now regarding Q 1 as the object for refraction at the 
second surface B, we note that p is to be replaced by 

QiB = QiO + OB = q' + ra, and - by ^? in the formula 

to M 
P-r __!* P 



So 



or 




or dividing by fiQ^'r^ 






86 GEOMETEICAL OPTICS 

On subtracting (a) 

J^L + J^I+l 
u.g TI r 



- < 



112 

Noting that r a = TV we see that ---- = , and by 

r 2 n r 2 

making P infinite we can find /"; similarly by making Q 
infinite we can find/'. "We have then 

i = ^v.2 and i.y^.l 

/ A* r a / /*' r 2 

and so we may write (5) as 

111 /' /" 

-- or = 1 



This tedious work might have been avoided had we been 
able to use the method of " Cardinal Points," but it is well 
to see the labour that is involved even in the simple case of 
a sphere, if we are ignorant of the better method that we 
shall shortly describe. It has been taken as an instance in 

f f" 
which our old formula 4. = 1 can be used if all the 

P <1 
distances involved are measured towards a certain point, in 

this case the centre of the sphere. The section on Cardinal 
Points will show that in any system, however complex, two 
points can be found such that, if the appropriate distances 

/' /" 
are measured towards them, the formula + = 1 will 

P 3. 

hold good. In the case of a sphere these two points are 
coincident. 

The Coddington lens, represented in Fig. 40, is a very 
convenient form of pocket magnifying glass, and has this 
important advantage over an ordinary convex lens the 
peripheral parts of the virtual image are as distinct as the 



NODAL POINTS 87 

central parts, provided that there is a central stop, so that 
none but centric rays traverse it. Usually a deep equatorial 
groove is ground in the lens, as indicated by the shaded part 
in the figure. In practice it is found that the central aperture 
must not be greater than a fifth of the focal length of the 
lens. The defects of the lens are : (1) the image is curved 
as the peripheral parts of the object are further away from 
than the central parts ; (2) the working distance is very short ; 
and (3) the field of view is limited ; as only those emergent 
pencils which can enter the pupil of the eye are effective ; it 
is advisable, therefore, to bring the lens as close to the eye as 
possible. 

The Stanhope lens is somewhat similar to the Coddington ; 
it is a short glass cylinder with its ends ground convex to an 




FIG. 40. 

unequal degree of curvature. The object is placed on the 
surface of lesser curvature, and the length of the cylinder is 
such that when the more convex surface is turned towards 
the eye, a distinct magnified image of the object is seen. 

Nodal Points of a Thick Lens. The following method is 
applicable only to lenses, and can, further, only be used 
when the initial and final media are the same. Should they 
be different, as in the case of the eye, it will be necessary to 
use the method described later (p. 91), which is perfectly 
general The only advantage of this method is its suitability 
in the case of thick lenses, if we forget the formulae for 
Cardinal Points. 

Given the thick lens, we first find the position of by 
(a), p. 83. We then find the position of the image of as 
viewed from the incident surface (A), and call it K', the 
first nodal point. We then find the position of the image of 
when viewed from the other surface (B), and call it K", 



88 GEOMETKICAL OPTICS 

the second nodal point. A warning must be given when 
dealing with menisci, as the point must always be con- 
sidered as situated in the glass, even when it is not, as in 
Fig. 37, or totally erroneous results will be obtained. Con- 
sequently the method must be regarded simply as a trick or 
device by which it can be mathematically proved that accu- 
rate results will be obtained. 

We will take the case of Fig. 38 as an example, where 
n = 3, r<i = 4, t = 2, and OA and OB we have found to be 
f- and f respectively. We are now considering the left- 
hand surface to be the incident surface A, so that light is 
supposed to be travelling from Q to P. The problem is the 
same as that on p. 57 ; we first find the image of formed 
by the surface A. As is in the glass, ju = 1*5 and // = 1, 
so 



Pfi" -f(-6) 36 

\T' A /-VM ft _ -tV * __ i ^ _ {_ __ _ _ _ 

r 2 - p -./"'" _| - 9 ~ -69 
Similarly for the surface B* 

,, 
/ - 

and /."-- .J; 

And as OB or p' 



12 
23 



2" 92 23 

The position of the nodal points is quite independent of 
the direction in which the light is travelling; it is only 
their names (first or second) that are changed. We have 
been considering light passing from Q to P, so K' refers to 
the left-hand surface. In the diagram the reverse condition 
is indicated ; the light is passing from P to Q, so that the 
right-hand surface is the incident surface, and the nodal 



SPHEKICAL ABEEEATION 



89 



point corresponding to it is called K', while K" refers to the 
emergent (left) surface. 

The mode of using these nodal points will be illustrated 
later, when we deal with the geometrical construction of the 
image of a complex system. 

Spherical Aberration. In Fig. 41 a beam of parallel rays 
is shown that encounters a double prism ; the more central 
rays SI and ST intersect at E after traversing the prism, 




FIG. 41. 




FIG. 42. 

while the more peripheral rays PJ and P'J' intersect at a 
more distant point, L. Clearly the aberration EL might be, 
obviated by bevelling the peripheral parts of the prism, so 
that the incident rays at J and J' would undergo a greater 
deviation. In a spherical lens (Fig. 42) this bevelling is 



90 GEOMETEICAL OPTICS 

carried out so that it acts like a prism, the strength of which 
is continually increasing from the axis to the periphery. 
Unfortunately, a spherical surface is not quite the right 
shape indeed, the bevelling has been carried too far, for the 
peripheral parts of an incident parallel beam intersect at a 
closer point than the more central parts. The diagram 
illustrates what is called spherical aberration (undercor- 
rected), while Fig. 41 will indicate what is called overcor- 
rection of spherical aberration. 

There is one obvious way of making the focus of such a 
lens more definite, i.e. by cutting off the aberrant peripheral 
rays with a stop, so that the focus is only formed by the 
intersection of the thin axial pencil. Another method which 
has certain advantages, as has been pointed out by Lord 
Eayleigh, is to block out the central rays and use only the 
peripheral ones. This, however, has been rarely used in 
practice. 

Just as with reflection at a mirror, when an oblique or 
eccentric incident pencil is considered, the refracted pencil is 
astigmatic, and presents the same focal lines with the same 
sphenoid shape between them. All this has been omitted in 
the diagram for simplicity. The short line on the axis 
between the focus of the peripheral and that of the more 
central rays may be regarded as the secondary focal line, 
while the intersection of the peripheral with the central rays 
indicates the position of the primary focal line sagittal to 
the plane of the paper. 

It may reasonably be asked, Why are lenses made of this 
erroneous shape? The answer is that it is impossible to 
mould glass of the right shape with any approach to accu- 
racy, and grinding by hand to any shape but spherical is a 
most laborious and difficult undertaking. 

If we confine our attention to thin oblique pencils, we 
see that they may be of two kinds : (1) an oblique eccentric 
pencil that is incident upon a peripheral or eccentric part of 
the lens ; and (2) an oblique centric pencil that traverses 
(either actually or virtually) the optical centre of the lens, 
as PQ in Fig. 38 or PQ in Fig. 37. The exact mathe- 



CAEDINAL POINTS 91 

inatical investigation of the form of pencil after (1) oblique 
eccentric refraction through a lens is most tedious and diffi- 
cult, and does not lead to any simple approximate expres- 
sion ; and, further, it is of little practical importance, except 
to certain instrument makers, so we must refer the scientific 
mathematician to more advanced treatises on this point. 
(2) Oblique centric refraction will be briefly treated in the 
Appendix (p. 121), as it is of considerable practical importance. 
"Cardinal Points. Gauss has shown us how to extend the 

f f" 
use of the formulae 4- = 1, etc., not only to thick 

lenses, but also to any refracting system however complex, 
formed of any number of media bounded by centred spherical 
surfaces. The only requisite is to find the position of two 
points called the Principal Points of the system under 
consideration. When the distance of the First Principal 
Point H' from the object P is denoted by PH', and the 
distance of the Second Principal Point H" from the object Q 

TjVTTf ~F"TT" 

is denoted by QH", we have ^w -f nTT ,, = 1 universally 



true, where F'H' is the distance of H' from the First Principal 
Focus F', and where F"H" is the distance of H" from the 
Second Principal Focus F" (Fig. 44). 

Before actually dealing with the problem, we will show 
exactly what it is that we want to find. Let HiA and H 2 B 
represent the principal planes of two thin concave lenses 
(Fig. 43). An incident ray parallel to the axis will, on 
traversing the concave lens at HI, be refracted in the 
direction F!^ as if it were proceeding from the second 
focus of the first lens FI ; on now meeting the second lens 
at H 2 , it will again be deviated in the direction H 2 R as if it 
were proceeding from K We wish to find a lens which shall 
have an equivalent action to these two lenses. It is clear 
from the diagram that a concave lens placed in the position 
KX, if its second focus be at NI, will have precisely the same 
effect on the incident ray SHiK as the combination of lenses 
had on SE^. The whole object of Gauss's method is to find 
the position of X, i.e. the situation of the equivalent lens, 



92 



GEOMETRICAL OPTICS 



and also its focal distance NX. In such a simple case as 
this, it is an easy problem to solve by purely geometrical 
means, but Gauss has shown us how to deal with any system, 
however complex. 




N A 

FIG. 43. 



The method used by Gauss is intricate and involves a 
considerable knowledge of mathematics, but as we know the 
result of his calculations, we shall be able to find the positions 
of H' and H" by easy algebra if we treat the question in the 
right way. In any algebraic problem it is necessary to pay 
the utmost attention to the algebraic statement of the problem, 
but after it is once correctly stated, think no more about 
the meaning of the future operations until you get your result. 
" Put it into the algebraic mill and turn the handle." The 
chief difficulty of all beginners in mathematics is that they 
try and think what each algebraic step means. This is quite 
unnecessary; thought is only required when stating the 
problem. 

We will take as an example the thick lens in Fig. 44, 
and use the following symbols to denote the various distances. 

The thickness of the lens AB = t, the distances of A 
from the first (fi') and second (fi") foci, due to the first 
refraction at A are f x 'A, =/i' and f/'A =/i", and similarly 
when f 2 ' and f a " denote the first and the second foci due to 
the second refraction at B, f a 'B =/ 2 ', and f 2 "B =/ 2 ". The 
values of these symbols can be determined in any given case 



CARDINAL POINTS 93 

by the well-known formulae of p. 55 ; they are indicated in 
the special case illustrated in the diagram by the letters 
below the line, whereas H' and H" that we have to find are 
shown above the line. We shall also use the symbols ti and 
h" to denote the distances H'A and H"B respectively. 

Now, if P denote any object (not illustrated in the 
diagram), PA or p will denote the distance of the surface A 




Stff 



from it, and we can easily find the position of the image Qi, 
due to the first refraction, for if q\ denote the distance QiA, 

g/i ' 
P - /i 

Kegarding now Q! as the object that forms the final image 
Q by refraction at the second surface B, we can as easily 
express QiB in terms of QB or q. For 

QlB = ^7? 

But QiB = QiA -f AB = qi 4- t, so we can eliminate qi by 
substituting for it our previous expression. We have then 



We have to find the positions of H' and H" so that the 
formula 

FH' F'H" 



shall give~a result that is identical with that given by (a). 

Now, ' PH' = PA -f- AH', or PA - H'A, i.e. p - ti ; 
similarly, QH" = QB - H"B = q - &", and if we denote 



94 GEOMETEICAL OPTICS 

FH' by F' and F"H" by F", this last formula can be 
written 



All, then, that we have to do is to make these two formulae 
(a) and (b) identical with each other. For (a) we shall 
write 



or 

pq(t +/r -/ a ') -yf/i'/t" + ft") + gffi'ft - ft 1 ) + *fW = o (O 

And for (b) we shall write 

F'q - F'b" -f F"p - F"h' = pq- ph" - qh' + k'h" 
or 



^ + ^') + ^'^" 4. w 4. jr%' = . (V) 

On comparing the coefficients of (a 1 ) and (5') we see that 
the two expressions will be identical if 



/ 4_ / " _ / ' fi'f*" + (/a" __ ^/i' fif* __ _ (/i/a" _ 
72 A" + ^"' 1^4-^' ~h'h" + F'h" + F"ti 

or calling this expression K, 



, 

A 

and if ^'A" + F'h" + F"h' = 



// /* / -f ' ' -f "-f ' 

.... AA ,. - -, if V $, if ^ = , and if F" = fl - 



these values are clearly consistent, and they are therefore 
the solutions required. 

In the case illustrated in the diagram (Fig. 44), the 
radius of the first surface, TI = 4, that of the second sur- 
face r 2 = 2, fjiQ = 1, u = 1/5, and t = 3. 



CARDINAL POINTS 95 

/'or ~ 
./i or , 



JJL fiQ 



5 




and K = t +ff -ft = 3 + 12 - 6 = 9 

fit -8x3 
^ = K= 9~ : 

j\, 7 

ft -4x3 



, fi't -8x3 M . 
.A=y 9 = -2|, 



If in this case we were to use the formula on p. 70, 
which neglects the thickness of the lens, we should have 



8 



or /" = - 8, and /' or -/" = 8, a result that is quite 
erroneous. However, on adding the correcting term 

- - , or on using the form 
- 



- 

r = r2 " ri " p) we get r = i 3 

.'./" or /' = 5j, which is correct. 

When two compound systems, A and B, are combined, 
A being the first system traversed by the incident 
light, t = Ha'H/, i.e. the distance of the first principal 
point of the second system (B) from the second principal 
point of the first system (A) ; whereas ti = H'H a ', or the 



96 GEOMETRICAL OPTICS 

distance of the first principal point of the complete system 
measured towards the first principal point of A. Simi- 
larly A" = H a "H 6 " and K= H a "H 6 ' +/"-/*', the sub- 
scripts a and b denoting the systems to which the letters 
refer. 

These formulae are a little difficult to remember, but 
when reference can be made to them they are easy to 
employ, and they are of universal application. We shall 
give three examples of their use, to illustrate their simplicity 
and value. 

Ex. (1). The ordinary form of eyepiece for a microscope 
(Huygenian) consists of two convex lenses, the distal one 
being called the field-glass, and the proximal one, to which 
the eye is applied, the eyeglass. It is found that a No. III. 
eyepiece, with an alleged magnification of eight diameters, 
cannot be used as a magnifying glass when held in the 
normal position before a microscope slide. Explain this, and 
show how it can be used as a magnifying glass. 

In all Huygenian eyepieces the field-glass (A) has a focal 
length three times that of the eyeglass (B), and the distance 
between them is one-half of the sum of the focal distances. 

The advantages obtained by this construction cannot be 
explained in this place, but it may be stated that the 
spherical aberration of the system is less when each lens has 
its second focus at the same point, and that the size (though 
not the position) of the image is achromatized when 

t = K/.' +/,') 

With a No. III. eyepiece f a ' = f ins., and // = f in. 
As we may regard the lenses of negligible thickness, we 
have t = | ins., 



and *<**+/."-/' = 4-4-4 = -f 



ET I Jajb 

= i^r- == ~ 



xs 



^=-*fj "* and ^"= -5 



HUYGENIAN EYEPIECE 97 

The position of the cardinal points of this Huygenian 
eyepiece is illustrated in Fig. 45, and it will be seen that the 
first principal focus (F') is 
within the eyepiece; and 
we know from p. 78 that 
the object must be placed 
not further off than F, so 
clearly the eyepiece will 
not act as a magnifying 
glass when used in this FIQ 

way. However, F" is situ- 
ated outside the lens, beyond B, so that if we reverse the 
eyepiece, using A as the eyeglass and putting the microscope 
slide in the position of F", it will form a very efficient mag- 
nifying glass. In such a position, when the eyepiece is 
reversed, F" and F' are simply interchanged. We have 
shown that the magnifying power of a lens is measured by 

^. In this case 4=10 ^--^ = 8. 

When a Huygenian eyepiece is used in a microscope, the 
image formed by the objective of the instrument is formed 
within the eyepiece at F', so that magnification results 
normally. 

If the two lenses of the eyepiece were placed in contact, 
the power of the combination would be 

r+p-f+f-l 

Ja Jb 

or the combination would then be equivalent to a lens of | in. 
focal length, while its magnifying power would be doubled, 
being 16 ; it would, however, manifest all the chromatic and 
aberrational errors that the Huygenian eyepiece in part corrects. 

Ex. (2). Suppose that the eyeglass in a No. III. Huy- 
genian eyepiece were replaced by a concave lens of equal 
strength : what would be the power of the combination, and 
what use could be made of the instrument ? 

Here/,' = - and K 01 t +/."-/.'- J_f + | = 

- 

00 



and as F' = -*& = m , the power = - 

A Jf 



98 



GEOMETKICAL OPTICS 



The power of the combination is therefore 0, i.e. all inci- 
dent parallel light emerges parallel, and it might be thought 
to be useless as an instrument; but a little consideration 
will show that it will act as a very efficient telescope 
indeed, this is the form of the Galilean telescope or opera 
glass. 

A very distant object, subtending a visual angle 9, will 
also make the angle f a "AD at the field-glass (Fig. 46) equal 




FIG. 46. 



to 8, and an image of the object would be formed in the 
second focal plane at f "D. But f a "D is also the situation of 
the first focal plane of B, consequently the incident parallel 
light which is converging towards D will, owing to the inter- 
position of the concave lens at B, proceed as a beam of 
parallel rays in the direction BD (the spaced and dotted 
lines). Similarly, the axial incident rays which, after 
traversing A, tend to converge towards f a " will, owing to the 
action of B, proceed as a parallel beam in the direction of the 
axis. Note, however, that the direction of the oblique pencil 
has been changed, as it now makes an angle 0' with the axis 
instead of 0. An eye, therefore, if adjusted for parallel rays, 
placed close to B will see an erect virtual image subtending 
the angle 0', and the magnification of the instrument will 
be 

tanj?' _f a "A_ -5 . -5 

tan 6 ~ 'B = "2 * 6 



CARDINAL POINTS OF THE EYE 99 

Ex. (3). We will now find the cardinal points of the 
human eye. According to Tscherning's most recent investi- 
gations, the lens of the eye has a focal length (0) of 51/34 
mm., and its principal points, indicated as H! and H a in 
Fig. 47, are so situated that HiAi = 2-308 mm., and H a A 2 
= 1*385 mm. The thickness of the lens is 3*9 mm., and it 
is placed 3*6 mm. behind the cornea (Ao). We shall con- 
sider the media to be of the same refractive index 1*3375, 
bounded by the cornea of radius - 7'8 mm. In the diagram 




FIG. 47. 



the incident light is presumed to be travelling from right to 
left, so we will regard this as the positive direction. 
Taking first the corneal refraction, we find 



and /"= -fif'= -30-91 mm. 

The distance t or 

AoHi = AoAi + A^ = 3-6 + 2-308 = 5*908 mm. 

and j^or t + /" - f = 5-908 - 30-91 - 51*34 = -76'342 

IT' A - T.' /'< 23>11 * 5-908 1 7QQ 
.-. H Ao, ^.e. h or - = -.- = - 1'789 mm. 



100 GEOMETRICAL OPTICS 



- -23-11x51-34 

1 OT F = -4 = ---- _ -- =15-64 mm. 



F'H" or J" - = ---- = - 20-79 nan. 



The distance of the second principal point from the cornea 
can be easily obtained, for 

H"A = H"H 2 + H 2 A 2 + A^ + A^o, 
or H"Ao = 3-973 + 1'385 - 3-9 - 3'6 = - 2-142 mm. 

The point H" is therefore only 0'353 mm. from H'. We 
shall not, then, introduce any appreciable error in consider- 
ing that they coincide in one point H, towards which both 
the focal distances are measured. 

Nodal Points of a Complex System. When we have found 
the cardinal points by the above method, it is a simple 
matter to find the nodal points. Take any point S in the 
first focal plane (Fig. 48), and through it draw S JiJ 2 parallel 
to the axis; join J 2 F". All light proceeding from S will 
emerge from the system in a direction parallel to J 2 F", e.g. 
SIiI 2 K. From S draw the ray SDiK' parallel to J 2 F", repre- 
senting such a ray, and from D 2 the point on the second 
principal plane corresponding to D b draw D 2 K"E parallel 
to J 2 F", cutting the axis in K". Then K' and K" are the 
first and second nodal points of the system. We shall see 
the use of them in the next section. Meanwhile we will 
devote a little attention to the diagram. We notice first that 
every ray incident on the first principal plane travels parallel 
to the axis until it meets the second principal plane ; this is 
a characteristic property of these planes, and as any object 
in one plane, when viewed from the other side, will be seen 
without any alteration in size, they are often called planes of 
unit magnification, or the Unit Planes. 

As the sides of the A F'SK' are parallel to the sides of 
the AH"J 2 F", and as F'S = H"J 2 , 

K'F = F"H"or I?". 

Also the A EF"K" = ADiJiS, as the sides are parallel, and 
as EF" = D 2 J 2 = D^!, 

/. K"F" = SJ X = F'H' or F' 



NODAL POINTS 



101 



It is also clear that K'K" = DiD 2 = H'H", 

and that K"H" = K'H' = F'H" + F'H' = F" + F. 




FIG. 48. 



We have found then all the cardinal points of the standard 
emmetropic eye which are given in the subjoined table. 



CARDINAL POINTS OP THE EMMETROPIC EYE. 



H'A . . 
H"A . . 


. . - 1-789 mm 
. . 2-142 mm. 


H'H" \ 
K'K" / 


. . . . 0-353 mm. 


K'A 
K"A . . 


. . 7-04 mm. 
. . - 7-39 mm. 


H'K' \ 
H"K"/ 


. . . . 5-25 mm. 


F'A . . 
F"A . . 


. . 13-75 mm. 
. . - 22-93 mm. 


F'H" \ 
K"F" / 


. . . . 15-54 mm. 






F"H" \ 
K'F' / 


. . . .-20-79 mm. 



In every system in which the initial and the final media 
have the same refractive index (i.e. when F"H" = H'F) the 
nodal points K',K" coincide with the principal points H',H". 
We see also that if we regard the principal points as coin- 
cident in the human eye in H, the nodal points coincide too 
in one point K, and the distance of this nodal point to the 



102 GEOMETKICAL OPTICS 

principal point, or KH, = F"H" + F'H' = - 20-79 + 15'54 
= 5 '25 mm. 

Geometrical Construction of the Image. To find the image 
of AB (Fig. 49), we first mark the position of the cardinal 
points on the axis. We then join AK', and draw K"# parallel 
to it. Then we can either draw AJ" parallel to the axis and 
J"F"a through F", or we can draw AFT through F', and IT'a 
parallel to the axis. In either case, a is the point of inter- 
section of the line with that drawn through the nodal point. 



. i . HT _ F^H' _ F'H' F' 

~ BA ~ F'B ~ F'H'-BH' " T r ^ 

1 la F"6 _ F'H" - 6H" _ F" - g 
o m H"J" ~ F"H" ~ F"H" F " 

On p. 75 it was shown that when thin lenses are con- 

sidered - = -, and we see from Fig. 49 that - = ^^-, so 
op o .bJi. 

that the nodal points of a complex system play the part of 
the optical centre (0) in a thin lens. The distances &K" and 
BK' are usually denoted by g" and g' in the books. 

In the simplified schematic eye described above, the nodal 
point K acts as if it were the centre of a convex spherical 
surface of radius 5*25 mm. that forms the boundary of a 
medium of refractive index 

F" 20-79 

~ F' r iF54~ 1338 

The tangent of the visual angle subtended at the nodal 
point of the eye (p. 77) by an object is the same as the 
tangent of the angle subtended by the retinal image at K. 
If this retinal image be i mm. in height, 



tan = 



It is often simpler, when the object is very distant, to con- 
sider an optical instrument and the eye as forming one 
complex system, as in that case tan is practically constant, 
and if K" denote the second Nodal Point of the system and 
K that of the unaided eye, it is clear that the magnification 
of the system is as 6K" : &K 



CONSTKUCTION OF THE IMAGE 



103 




104 GEOMETKICAL OPTICS 

As an example, we may explain the action of a Baden 
Powell lens, which is used as a handy pocket opera glass. 
It is simply a convex lens of weak power which is held at 
arm's length, and distant objects are viewed through it; 
these then appear larger to the eye, or, when less correctly 
expressed, they are said to seem nearer. 

We will suppose that the glass is + 0*5D where /' 
= 2000 mm., and that it is held at a distance of 500 mm., 
or about 20 inches, from the eye. 

On considering the lens and the eye as one complex 
system, and finding the cardinal points, we get the following 
values : 

K= t +f _ F' = 500 - 2000 - 15-54 = -1515-54 

,, ft 2000 x 500 
" * r K = -1515-54- = ~ 
, F"t -20-79 x 500 = 6-859 mm. 
h Or ^r = -1515-54 
, _ -/'J __ -2000x15-54 

K -1515-54 ~ * 

p ,,^f"F" -2000(-20-79)^ 
~K~ -1515-54 

It is clear, then, that the first principal point and the 
first focus are both more than 100 mm. behind the eye. 
This is of no importance in the problem now before us ; but 
the second principal focus must be situated on the retina if 
the object is to be seen distinctly, so we must find whether 
this is so. 

F" is situated 27'436 mm. behind H", which is 6*859 mm. 
in front of the eye, consequently F" is 20*577 mm. behind 
the principal plane of the eye. As in the standard eye F" is 
20*79 mm. behind its principal plane, we see that, unless the 
eye be 0*213 mm. too short, a sharp image will not be formed 
on the retina. 

Now, few eyes are of exactly the standard length, and 
we will suppose that the eye considered is too short by 
at least this small amount (less than 0-75D of hyperme- 
tropia), so that 6K" = F"K" = -F'. 



GRAPHIC METHOD 105 

The magnification (M) will be as F"K" : F"K 

F"K" -F' 20-51 
= F'K = ^F' = T5^54~ 13 

Consequently, the distant object will appear larger by 
nearly a third. If the eye be myopic or too long, the Baden 
Powell lens is useless without some contrivance to make the 
image definite. The simplest effective contrivance is a card 
with a pinhole in it held close to the eye. In this way the 
circles of confusion on the retina are made much smaller, so 
that the image .may be regarded as sharply defined. 

Graphic Method for Cardinal Points. Professor Sampson 
has devised a most ingenious method of finding the cardinal 
points of a thick lens by a graphic method. It is an exten- 
sion of the graphic methods which we have frequently illus- 
trated in the preceding sections. 

The diagram (Fig. 50) shows this method applied to the 
case of the meniscus we discussed on p. 95. The thickness 
of the lens AB is 3, so we measure AB in the positive direc- 
tion (upwards) 3 units. As A indicates the first surface, we 
mark off fi'A in the negative direction to represent - 8 units, 
and fi"A in the positive direction to represent 12 units. 
Similarly, dealing with the second surface B, we make f 2 'B 
equal to +6 units, and f 2 "B equal to -4 units. The paral- 
lelogram referring to the first surface (A) is completed at KI, 
and that referring to B is completed at K 2 . 

Join KxK 2 , and produce to meet the horizontal line 
through B at H", cutting the horizontal line through A at H'. 

Then H'A is ti, the distance of the first surface from H', 
and H"B is /*/', the distance of the second surface from H". In 
this case they are both measured from right to left, so they 
are both negative : H'A = - 2f , and H"B = - 1. 

Join KI and f 2 ', and produce to meet the horizontal line 
through A in F'; and join K 2 fi", and produce to meet the 
horizontal line through B in F". 

Then F'H' is positive and is found to measure 5 J units, 
while F"H" is found to measure - 5 J units. It will be found 
that F'H' and F"H" correctly indicate the two principal 
focal distances of the thick lens illustrated in Fig. 44. 



106 



GEOMETRICAL OPTICS 



The method is of delightful simplicity and presents no 
difficulty whatever, if due attention be paid to the directions 
of the positive and negative measurements, and if it be noted 
that the points fi" and f 2 ' must be in the same straight line 



f A 


\ 

1 

\ 


7 


\ 


\ 


' 




\ 





/ 




\ 


\ 


I 




\ 


\ 


1 




Si 


\ A 2 






\ ' x 






\/ \ 






/\ 






/ X - X 






\ 






/ > \ 






/ 'S;\ 




s# 


/ ^^ 


IS 



FIG. 50. 



as AB, and that K x is joined to f 2 ' (subscript 2), while K 2 is 
joined to fi" (subscript 1). 

To show the generality of the construction, it will be 
sufficient here to notice that 



f^'fa' = fi"A + AB - f 2 'B = t +/i" -&+*, 
and that from the similar triangles in the figure 

FA f 2 f A or FH^+ H[A = fa f B_-_AB 



or 



, .. . F'B fj"B 

Similarly, ~ fl = T^nr, 

K 2 i 2 ii i2 



LENSES 107 

F'H" + H"B ri "A + AB 



or 



or - (Cf.p.94) 

/2 

(The above is not given as a proof of the construction, for 
clearly FA = FTK' + K'A, and F'B = F'K" + K"B, or indeed 
the sum of any other lines with the same end points. It 
will be found that if the final medium has a different re- 
fractive index from that of the initial medium, the points 
marked H' and H" really denote the two nodal points K' 
and K" of the system. As from p. 101 we know that when 
F" = F', the nodal points coincide with the principal 
points, we may regard the points H' and H" determined by 
this construction as always denoting the nodal points.) 



QUESTIONS. 

(1) The focal length of a convex lens is 6 ins. ; an object is placed 
36 ins. from it. What is the relative size of the image, and where is 
it formed ? 

(2) Show that/'/" =(/' - p)(f" -q) in all cases. 

(3) The radius of curvature of the first surface of a thin converging 
lens is 6 ins. If its focal length be 10 ins., and if the index of refrac- 
tion be 1-52, what is the radius of curvature of the other surface ? 
What would its focal length be when placed in a tank of water 0* = ) ? 

(4) A convex lens of focal length i in. is used as a magnifying 
glass. The nearest point of distinct vision is 10 ins. from the nodal 
point of the eye. Find the magnifying power (i.) when the lens is in. 
from the nodal point, (ii.) when it is 1^ in. from the nodal point ; and 
(iii.) when the object is i in. from the lens, and the eye is adapted for 
distance. 

(5) The cardinal points of the following thick biconvex lens are 
required where r r = 4, r 2 = 2, t 3, and /x = T5. Use the graphic 
method, and test your results by the algebraic formulae for the position 
of the cardinal points. 



APPENDIX 



THE following notes are intended for those who require some 
further knowledge of the subject which the previous elementary 
treatment rendered impossible. At the same tune care has been 
taken to include nothing which has no practical application ; 
academic points of merely mathematical interest have been rigidly 
excluded. 

Oblique Reflection. Distances of the Focal Lines. We will 
now give the proof of the formula quoted on p. 33, denoting 




FIG. 51. 

SP by u and F X P and F 2 P by v l and v 2 , the distances of the focal 
lines due to the thin incident pencil PSQ (Fig. 51). 

PCQ = OCQ - OOP = CSQ + SQC - (CSP + SPG) 

= CSQ - CSP + JSQF, - JSPF, 
2PCQ = 2PSQ + SQFi - SP^ / (a) 



110 APPENDIX 

But owing to the equality of the angles formed by the intersect- 
ing lines SP and F X Q, PSQ + SQF, = PFiQ + SPF W and on 
substituting this expression in (a) we get 

2PCQ = PSQ + PF,Q ..... (J) 

Now, as the formula that we wish to find is only true for thin 
pencils PSQ and PF X Q, we may substitute the chord PQ for the 
arc PQ, and regard PSQ and PF X Q as triangles; moreover, 
CQS or FjQC = <. 



* r>Qn Sn sn 

Then m A PSQ, gp - ETQS " E{90 3 +" CQS) 

PQ = sin PSQ 
'* SP ~ : cos < 

andinAPFO PQ _ sin PF,Q _ sin PF,Q 

m & W f* F X P ~ sin PQF, ~ sin ( 

PQ sin PF t Q 
'''Ff~~ cos<j> 

therefore in the limit when PSQ and PF X Q are very small 

= . (c) 



Note that as PQ is measured from below upwards, but SP 
from right to left, the angles PSQ and PQS are measured in 
reverse directions ; a similar precaution must be exercised in 
dealing with the angles PFjQ and PQFj. 

On substituting the expressions in (c) for those in (b), we can 
write 

PQ COS< , po - ^ 



. 

u v- r cos 



Again, since A F 2 PC + A GPS = 

sin <j> -f- ru sin <^> = V#JL sin 






or, on dividing throughout by \v#u sin ^>, 

1 1 __ sin 2< __ 2 sin < cos rf> __ 2 cos < 
u v 2 ~~ r sin <^> "~ r sin < r 

The reader may be inclined to think that the consideration 
of such very thin pencils is of little practical use, but it must be 



CAMERA LUCIDA 



111 




remembered that only very thin pencils can enter the pupil of 
the eye, so that only thin pencils need be considered, if the 
image formed by the mirror is viewed directly by the eye. 

Camera Lucida. The Wollaston prism shown in Fig. 52 is 
the usual form of camera lucida that is used in sketching objects 
from nature. It is a glass 
prism that presents four 
angles, one of which is 90, 
the opposite angle 135, and 
the remaining two angles are 
each 67J. Light from the 
object enters one of the faces 
normally, and is transmitted 
without deviation to the 
second face as SI; at I, 
however, its angle of inci- 
dence is 67^, much greater 
than the critical angle for 
glass, so that it is totally 
reflected at I as IR ; again 
at R it undergoes total reflec- 
tion; and finally it emerges Q 
normally to the upper sur- FlQ 52> 
face towards the eye at E. 

The eye will therefore see a virtual image of the object pro- 
jected downwards in the direction EQ. In practice the eye 
is placed over the edge near T, so that one half of the 
pupil receives the light from the prism, while the other half is 
viewing the sketching-block and pencil placed below in the 
neighbourhood of Q. It is most important that the image of 
the object should be accurately projected on to the plane of the 
sketching-block, so the upper border at T is ground concave so 
that it will have the effect of a 4D lens. In this way the 
image is projected about 10 ins. from the prism, which will be 
a convenient distance for the sketch to be made. The image 
seen is erect, as there are two reflections ; had there been only 
one reflection the image would have been upside down. 

Oblique Refraction at Plane Surface. Focal Lines. We 
will now consider the refraction of a thin oblique pencil at a 
plane surface. If POQ (Fig. 53) be an incident oblique pencil 



112 



APPENDIX 



originating from the source in a rare medium and refracted at 
PQ in the direction RB', the primary and secondary (virtual) 
focal lines will be formed at F x and F 2 . As the pencil considered 
is very thin, we may substitute the angles POQ and PF X Q for 
sin POQ and sin PI^Q, and we may disregard the difference 
between the angles AOP and AOQ and consider that they are 





FIG. 53. 

each of them equal to <f>, and similarly we may consider F 2 as a 
point, and consider that AF 2 P = AF 2 Q = <'. 

If we denote OP by u, Ff by v lt and F 2 P by v& we have 



in A POQ, 
and in 



PQ __ sin POQ __ sin POQ 
OP "~ sin PQO "" coTAOQ 

PQ 



so 



sn 



POQ = ^ cos $ and PF X Q = ^ cos 



Now, 



sn 



sn 



cos 



cos <f> 



(a) 



and we wish to find from Fig. 53 an expression for -,, or the 



,. ... - 

limiting value of 



A 



, i.e. 



POQ 



FOCAL LINES 113 

When, therefore, POQ and PF t Q become infinitesimal, 

PQ 
d<f> u v l cos < p.' cos < 

- 



,AX 

(A) 



AP AP 

And since sin < = and sm 9 



/ sin <j> _ F 2 P 
r sln~' - OP 



(B) 



The refracted pencil will be astigmatic, and a sphenoid will 
be formed between the secondary and primary focal lines (F 2 and 
Fj) exactly like that described on p. 65, when oblique refraction 
at a concave surface was considered from a source of light whose 
distance u was less than /u, (FjP). A blurred image of the 
point will be formed at the position of the "circle of least 
confusion " between F 2 and F l (as represented by D in Fig. 56, 
where, however, F,>P is greater than FjP). 

Oblique Refraction through a Plate. Let t denote the thick- 
ness of the plate, and let I denote the length of the path NM of 
the thin pencil through the plate in Fig. 19. For the first 

refraction we have v- u . ^ and v 2 = ~u by (A) and (B). 
^o cos j < [j, 

For the second refraction at M the angle of incidence is <', and 
that of emergence is <, while the relative index is , , so that on 

replacing u by i;, + I or v z + I in the respective equations, and 
making the other appropriate substitutions, we get 

. (A') 
..... (B') 



The distance between the two foci, F 2 - V, or^l 1 - 




fj. \ COS <p 



114 APPENDIX 

may be taken as a measure of the indistinctness of the image. 
However oblique the pencil the distance separating the two foci 

cannot exceed ~, I ; for direct pencils of course < = 0, and the 

two foci are coincident at one and the same point distant u + tl 

from the distal surface of the plate. If preferred we may sub- 
stitute t sec <' for I, but the form given above is easier to 
remember. 

When a small object S is seen obliquely through a glass plate 
(Fig. 54) the following points may be observed : 

(1) The image S' is 
blurred, as it is represented 
by the" circle of least con- 
fusion" between the two 
focal lines. 

(2) The upward dis- 
placement is greater than 
when the object is viewed 
normally, for in that case, 
as we found on p. 40, the 
upward displacement is 

^ t, or one-third the thickness of the glass plate. 

(3) There is in addition a lateral displacement. 

Refraction at Spherical Surface. Focal Lines. The length 
of Vi is determined in this way (Fig. 55). When the pencil is 
thin, CPO or CQO may be represented by <, and CPF 2 or CQF 




sn s sn 

a A 



_ 

sin (W^OQC) cos CQO 

Similarly, in A F a PQ, 

PQ _ sin PFjQ _ sin PF X Q _ sin 



FjP ~~ sin PQF, sin (90 - F a QC) cos CQF 2 
Therefore in the limit when POQ and PFjQ are very small, 
PQ C , PF lQ = PQ' and POQ = * 



FOCAL LINES 

Now, /AQ sin <f> = p.' sin <j>, 



115 



cos 



cos 



As we wish to find from the figure an expression for --, , or 
the limiting value of -T~?>J we must no longer disregard the differ- 



ence between CQO and CPO and that between 
Since the angles at the intersection L are equal, 



i and CPF,. 




or 



or 



FIG. 55. 

POL + LPO = LCQ + CQL 
POQ + < 



And similarly, since the angles at K are equal, 
PFjK + KPFj = KCQ + CQK 



4- 



. A<A _ 
" 



or in the limit 
But 



- PCQ 

S <> 1 




s $ _ 1\ 
v f) 



116 APPENDIX 

fl' COS 2 <' fl' COS <f> _ COS 2 <f> COS 



r 

\ 

- cos <f> ) 
V 

sn < cos <' cos <j> sin <' 

r sin <' 
_ S inQ- <#>') 

OI - . - r t ~ > I .tV J 

/*, w r sin < 

Again, since 

ACPO = ACPF 2 + AF 2 PO 
Jrw sin < = Jr^ sin </>' + Jv 2 w sin (<#> <') 

or on dividing throughout by fyruv z sin <^>' 

sin < 1^1 sin > <' 
' ~ 



sin <f> V 2 ~ u r sin 



These formulae (A) and (B) are only true for extremely thin 
pencils, but it is only such that can enter the pupil of the eye, so 
that they are applicable in all the optical questions relating to 
oblique vision in the human eye. They will also be used in the 
next section when we deal with thin oblique centric pencils 
traversing a lens. 

It is evident from formula (A) and (B) that 

-- ( 1 - sin A') - 1 - rin '* = JL _ 1 
MtV 

or !_ 



Ho V l U 

fi' . V 2 sin 2 

- (v. - Vl ) = ^ 



When yu, has the value 1, this expression takes the form 



sn 



and we see that when ^ 2 and ^ x carry the same sign, i.e. when 
the two focal lines are on the same side of the refracting surface, 



^! as w= 



CIRCLE OF LEAST CONFUSION 117 

On considering the caustic curve formed by a wide divergent 
pencil issuing from a point source, it is clear that the cusp of the 
caustic points away from the spherical surface when v z > v lt but 
that it points towards the surface when v 2 < Vj. No caustic is 
formed when v 2 = V 1 ; this can only occur when < = 0, or when 
u = fj.v l ; the refracting surface must then be concave, for if 
< = 0, u = r t and if u = pv^ as v must be positive, it must be 
greater or less than u, according as /u, is less or greater than 1. 

Circle of Least Confusion. It has been shown that if the 
surface be not aplanatic for the source, whenever the incident 
pencil is oblique the refracted or reflected pencil is astigmatic, 
crossing itself in two focal lines and tracing out a sphenoid sur- 
face between them. It is required to find the position and size 
of the smallest cross-section of this sphenoid surface represented 
by DK in Fig. 56, which is practically a repetition of Fig. 15. 
The surface (ab) at P is that part of the mirror or lens which 
gives rise to the astigmatic pencil that has its focal lines at ^ 
and F 2 . For geometrical convenience squares have been described 
on the radius (k) of the circle of least confusion, and on the 
radius R of the effective aperture of the receiving instrument. 
Now, DP (or x) and k, i.e. the distance and side of the small 
square at D, can be easily expressed in terms of a and b. Note 
that when comparing the vertical sides of the rectangles at P and 
D, while a is measured upwards k is measured downwards, so 
that they carry opposite signs. 

a=~k or ~F '''^ = X ~1T ' ' (V) 
F 2 P_F 2 D . v*_v z -x 
~b"~^k~ ' 0~ k (2) 



On eliminating k from (!') and (2') we obtain 

x v l -.Vz x 
a = - 



which, on dividing by x, reduces to 



118 APPENDIX 

In practice, however, we have more often to consider the 
cases in which a real image is formed at the back of an eye or at 
the back of a photographic camera. It is, therefore, much more 
convenient to express the distance (x) and the radius (k) of the 
circle of least confusion in terms of the distance (s) of the 
receiving instrument and the radius (R) of the stop used with it. 

Let H' represent the first principal plane of the receiving 
instrument (distant H'P or s), which only allows a pencil of 
radius R to be transmitted through it ; if the receiving instru- 
ment be an eye, R represents the radius of an equivalent pupil 
placed in the first principal plane of the eye. 

By similar triangles we have 

FT!! FT!) DF, . #1 s x v\ /, \ 

or 



R -k k R k 

F 2 H F 2 D v 2 - s # 2 - x 



and ~BT ~T ' ' ~IT T 

i\ -f #2 2s v 2 



R(v } 2 - v^ 

or k = V^ - r . . (a) 

- - 



Again, from (1) and (2) we also see that 

k _ x v l __ v 2 x 
R~~ Vi s ~" # 2 s 

or x(v z - s) - vj)* 4- ^s = v v v z v<,s x(v v - s) 

.-. X(V-L + v z 2s) = 2viV^ s(vi -f v 2 ) 
2^-sfa + t,) 
Vi 4- ^2 2s 

These two equations () and (b), determining the radius and 
the distance of the circle of least confusion of an oblique pencil, 
are always true whether reflection or refraction are considered. 

If the receiving instrument be an eye, and the lens at P be a 
spectacle glass worn in its appropriate place (the first focal plane 
of the eye), a simple expression can be obtained for the radius r 
of the retinal confusion circle formed by the retinal image of this 
confusion circle at D. 

For r F> 

~ 



CIKCLE OF LEAST CONFUSION 119 




120 



APPENDIX 



But as the lens is placed in the first focal plane, PH' = F', 
so 8 = -jT, and F' -p = PH' - DH' = PD = -x; 

sR(v z - flQ 



_ = 

= ~~ 



- s(v! 4- 






Least Circle of Aberration. These formulas are of no use 
when we are considering the spherical aberration of a lens 
used with full aperture, as in Fig. 42. Let y (Fig. 57) denote 
R 





FIG. 57. 

the semi-aperture of a convex lens the focus of which is at F", and 
let BTL denote an extreme ray ; it is clear that if a screen were 
placed at F", there would be a bright point at F" which would be 
surrounded by a halo of light extending to L and L'. The Jine 
F"L is called the Lateral Aberration (I) of the extreme ray RTL, 
while F"T or a is called the Longitudinal Aberration of RTL 
which cuts the axis at an angle H"TR or \jr. 

Clearly a = F'H" - TH" = f " - y cot ^, 

and I = a tan \j/. 

The point E represents the position of the first focal line (/j), 
and the point T that of the second focal line (/ 2 ) of the extreme 
ray RTL, and the curve EF"E' represents the Caustic Curve 
formed by the lens when its effective aperture is R'R. It is 
obvious that the light is concentrated over the smallest area at D, 
which marks the site of the Least Circle of Aberration, its 
radius DK or k is determined by the point where the extreme 
ray RTL cuts the caustic E'F". It is usually stated that 
F"D = a and that k = l 



THIN OBLIQUE CENTRIC PENCILS 121 

This is only a very rough approximation, as the position of D 
can only be determined by tracing the caustic, 1 which may be of 
almost any shape, some being long and narrow, while others are 
short and stumpy, so that no general expression can be given for 
the position of D. If in any case the position of D is given, of 
course Tc = TD tan \j/. 

Thin Lens. Thin Oblique Centric Pencils. It has been 
already demonstrated that all oblique pencils which traverse the 
optical centre of a lens emerge at an angle equal to that of 
incidence (<j>) at the first surface, and that the angle of incidence 
at the second surface is equal to the angle of refraction (<') at 
the first surface. This fact is universally true, however different 
the radii of curvatures of the respective surface may be. 

Fig. 58 gives in a simple diagrammatic way the results of the 
refraction of a thin oblique pencil PO that traverses the centre 
O of a thin concave lens represented by the plane HO. 




FIG. 58. 

By the method of p. 116 we find the positions of p l and p. Z) 
the primary and secondary focal lines due to refraction at the 
first surface. Then, regarding p l as the virtual object for the 
second surface, we find g 1 as the final primary focal line due to 
this oblique centric refraction through the lens, and similarly 
q represents the final secondary focal line. 

1 A paper of mine in the Proceedings of the Optical Convention, 1912, 
describes a method of tracing Caustic Curves. 

I 2 



122 APPENDIX 

Hence, if r^ and r 2 represent the radii of the first and second 
surfaces respectively, and if PO = U, pf> = v lt and p 2 O = # 2 , we 
have for the first refraction 

p! cos 2 $ _ cos 2 <ft __ // 1 __ sin (< <') . ^ 

U "~ A 2> 2 U " T sin <>' 



For the second refraction <' is the angle of incidence, < is 
that of refraction, and ? is the relative index, so if qf> = V^, 
and q 2 O = Vjj, we have 

x cos 2 < cos sin ' - <> 







Vi jw 2 v z r. 2 sn </> 
cos 2 1 - sin <> ~ 



sn 



On adding (a) we obtain 

1 1 sin - 



o, 

cos 



This formula is, of course, only true for thin centric pencils ; it 
is of no use when the whole aperture of a lens is employed to 
form an image, but only when a diaphragm with a small central 
perforation is used with the lens. It is, therefore, applicable to 
the case of a camera lens when a small " stop " is used to make 
the image more sharp, but perhaps the most important professional 
use of it is in the case of spectacles. The pupil of the eye then 
limits the width of the effective pencil, and when the visual lines 
traverse the centre of the spectacles an application of the formula 
gives rigidly accurate results. If the wearer gazes through 
eccentric portions of his glasses, the results are only approximately 
true, as then an investigation into the problem of oblique 
eccentric refraction would become necessary. However, for all 
practical purposes the simple formula we have obtained will 
suffice. 

Short-sighted persons are often seen wearing their pince-nez 
tilted on their nose, and they declare that they see better with 
them in this position than when they are placed vertically. It 
will always be found in such cases that the myope is astigmatic, 
and that his pince-nez do not correct his astigmatism. He is, in 



TILTED LENSES 123 

fact, making use of this property of tilted lenses to obtain an 
astigmatic pencil which shall correct his error. I personally very 
frequently order tilted spectacles to poor patients to whom the 
expense of correct sphero-cylindrical lenses is prohibitive. Tilted 
lenses are of use when the power of the spherical part of the 
lens is high, and when only a weak cylindrical effect is required 
in vertical meridian, i.e. when the plane axis of the cylinder is 
horizontal ; they have this incidental advantage, that they are 
lighter to wear. 

We will take an example to illustrate this application of the 
formula, as it will show the most convenient way in which such 
questions may be treated. 

Ex. A patient wears his distance pince-nez ( 10D) tilted 
30 from the vertical plane. Calculate the effect of this dis- 
placement. 

As the object is presumed to be at a great distance, the 

terms involving - vanish, and remembering that 



when distances are measured in centimetres, we may write the 
equation (A) as 

D sin <-<*>' 



cos 2 



- . -- 
I sin 



where D stands for 10 dioptres, D l for the dioptric power of 
the glass in the vertical meridian, and D 2 the dioptric power of 
the glass in the horizontal meridian. 

When < = 30 and /A = 1'5, D 2 is found to be very nearly 
10-95 dioptres. Consequently, 



that is, the 10D glass with this tilt will act as if a cylindrical 
glass of power 3*65D, with the plane axis of^the cylinder hori- 
zontal, had been added to a spherical lens of 10-95D. If we 
denote the cylindrical glass by D c we may find its power more 
shortly thus 



:.-D, = K- 10-95) =5-3-65D. 



124 APPENDIX 

Decentration of Lenses. It is sometimes said that the 
refracting power of a lens is measured by the deviation that it 
induces on a ray of incident light. On referring to Fig. 59 this 




FIG. 59. 

statement is seen to be untrue, or at any rate inadequate. The 
First Focus of the convex lens is indicated by F', and the ray 
F'H on emerging from the lens pursues the direction HS ; its 
direction therefore undergoes a deviation F'HS' or HF'O, and it 
is clear that the deviation HF'O depends upon the height of OH, 
i.e. it depends upon which incident ray we choose to consider. It 
will be noticed that the angle HF'O is negative, being measured 

TTO / 
in the clockwise direction, so tan HF'O = WQ or ji where HO is 

denoted by /. The lens, considered only with reference to the 
ray F'H has the same effect on it as a prism with its edge 
upwards ; indeed, a lens may be regarded as a prism the strength 
of which is continually increasing as one passes from its 
centre O. 

Now, if an eye were at S, it would see the image of an object 
at F 7 displaced to a great distance in the direction of S'. The 
lens would be said to be " decentred " HO millimetres downwards 
(with respect to the eye). If the lower part of the lens were cut 
away it would be called a Prismosphere, as it would act precisely 
like a spherical lens which had been bisected equatorially with a 
prism inserted between the two halves. Recently a great deal of 
attention has been directed to the properties and applications of 
decentred lenses or prismospheres, so that it may not be out 
of place to give the very simple formula which connects the 
decentration I of a lens with the deviation it induces. 

Prism Dioptres, Centrads. The Prism Dioptre is a unit 
proposed by Prentice, and is that angle whose tangent is 0*01 and 
is symbolized by delta (A). Consequently, with a -f 1 D lens of 



PRISM DIOPTRES 125 

which /' = 1 m. or 100 cm., when OH (Fig. 57) is 1 cm. a 
deviation of lA is produced, for 



In the prescription of spectacles the decentration, HO or I, is 
always measured in millimetres, so that we have the simple 

formula N = -TQ A, where N denotes the number of prism 

dioptres. 

In order to obtain all the information possible from this 
formula we must make some special convention about signs. A 
prism, for instance, with its edge towards the right before the 
right eye would cause divergence of the eyes, but if so set before 
the left eye, it would cause convergence of the eyes. If, however, 
we agree to call prisms with their edges upwards or outwards 
(from the nose of the patient) negative, we shall get consistent 
results. Further, if we regard decentration downwards and 
inwards as also negative, the above simple formula will give us 
full and complete information about the prismatic equivalent of 
all decentred lenses. This convention about signs is a little 
difficult to remember ; the memoria technica that I use is 
Ztecentratiow, down and in, negative; Prisms opposite. For 
instance, what effect will a 5D lens have if decentred 4 mm. 
upwards ? This means what effect will a spectacle lens have if 
its optical centre is displaced 4 mm. above the pupil of the eye ? 

Here I = +4, it is positive because the decentration is 
upwards, and 



The decentration is therefore equivalent to a prism of 2 prism 
dioptres with its edge set in the negative direction, and from the 
memoria technica we know that must be either upwards or 
outwards, and from the conditions of the question we know that 
it must be upwards. The effect of this decentred lens will be the 
same as that of a normally centred 5D combined with a prism 
of 2 A set edge upwards. 

There are, however, some objections to the prism dioptre as 
a unit. It is not subject to the ordinary rules of arithmetic ; 



126 APPENDIX 

for instance, 2A -f 3A are not exactly equivalent to 5A. This 
difficulty has been entirely obviated by Dennett's unit of the 
centrad, which is the hundredth part of a radian, and is denoted 
by a reversed delta (V). Each unit is practically 34 '377', but 
the multiples of a centrad are appreciably greater than the mul- 
tiples of a prism dioptre ; and this is even of an advantage in 

using the formula N = ^V, owing to spherical aberration. It 

has been adopted as the official unit by the American Ophthal- 
mological Society, and before long I hope it will be universally 
used in this country. When the centrad is used, 3V -f- 4V are 
exactly equivalent to 7V, and 10V is ten times the strength 
of IV. 

What is the effect of a +2-5D lens decentred 4 mm. in? 

ID (-4)(2-5) 
~ 10 V 10 

The effect will be that of a normally centred +2-5D com- 
bined with a prism of one centrad set edge outwards (i.e. it will 
be abducting in function). 



FORMULA 127 



FORMULA FOR REFERENCE. 
A. Universally true 
(a) For thin centric pencils 



For oblique pencils when ^ and v. 2 carry the same sign 



=v as w 



(c) Circle of least confusion 



f i + *> 2 - 2s 
- sfa + 



i 4- Vz - 2s 

B. Refraction at a Spherical Surface 
When p Q is the index of the medium in which the source of 
light lies, and // is the index of the refracting medium 



- Mo Mo 
ECCENTRIC PENCILS 
2 



sn < - 



/AO u fv 2 M r sn 

C. Lenses 



MAGNIFICATION 

M 

B 



( 

When KJ = J = rf + q, (i) becomes ^TT*. (u) becomes 



128 APPENDIX 

. CARDINAL POINTS 



When =+//' -/ a ', 
f-t /*"/ ff fiifii 

ii /I* in _/ l _ "Vi/g TJH /1/2 

h= K> * ~K' F ~-~K~' F = ~K' 
In complex systems, t = H a "H 6 ', V = H'H a ', h" = H"H 6 ". 
DIOPTRES 

j QQ 

When f given in centimetres D = -jr\ when/' in millimetres, 



DECENTRATION AND CENTRADS 



THIN LENSES, OBLIQUE CENTRIC PENCILS 



(ii) When incident rays are parallel 
D sin Q- 4Q n. 

D * -~jT^i ' ~~8i^' ' DC 

D. Prisms 

D = i/f - <#> - A. 

MINIMUM DEVIATION 

D = 2,/r - A. 



sin }A 



If A small, D = p - 1A. 

In order to convert the formulae (B) into those suitable for a 
plane surface, put r = oo . To convert them into the correspond- 
ing formulae for reflection, put /u, = 1 and <j>' = <j>. 



ANSWERS 



CHAPTEB I (p. 10). 

(1) 45 feet. (2) 70 feet. 

(8) The intensity of the electric light is 900 times that of the 
gaslight. 

(4) The shadow cast by the gas-flame will be approximately 8 
times more intense than that cast by the electric light. 

CHAPTEE II (p. 17). 

(1) 1 foot, 5 feet, 7 feet. (2) 7 images. 

(3) One image. If ECB < 60, two images would be seen. For 
PCB = 120 since AGP = 15 and ACB = 135. Produce BC to c, 
and draw Pp 6 perpendicular to BC produced, cutting it at c ; make cp b 
= PC ; then the angle p b Cc = 60. Consequently, the line p 6 E will 
only cut the mirror when ECB < 60. 

CHAPTEB III (p. 33). 

(1) The image is 15 cm. in height, it is virtual and erect, and it is 
formed 16 cm. behind the mirror. 

(2) (i.) 24 inches in front of the mirror, 
(ii.) 8 inches behind the mirror. 

(3) r = - 6 inches ; the mirror is convex. 

(4) p = 6^ inches, - = - 11 ; the image is inverted. 



CHAPTEB IV (p. 48). 

(2) M = 1-54. (3) n = V2. 

(5) Two prisms of minimum deviations 1 44' and 2 respectively, 
if properly placed, would correct the deviation. The weaker prism 
should be placed edge outwards before the left eye, while the 2 prism 
should be placed edge outwards and upwards before the right eye, so 
that the base apex line makes with the horizontal line an angle of 30. 

129 



130 ANSWEKS 



CHAPTEE V (p. 67). 

(1) An inverted image mm. high would be formed 22'2 mm. 
from the cornea, i.e. 2-2 mm. behind the second principal focus. 



150 _ 1 

~ io' lc ~ 



I a _ -0-6 ._ 10 
I c ~ -0-6 ~ 9* 

(3) A real image will be formed at the distance of the diameter of 
the sphere from the unsilvered side. 



CHAPTER VI (p. 107). 
= -7-2 inches. 




=-' /./'/"=CT - 

(3) r 2 39 inches ; in water /'= 37 inches. 

(4) (i) M = 48-5, (ii) M = 43-5, (iii) M = 50. 

(5) h' or H'A = -1-6, h" or H"B = 0'8, FH' = 3-2, F"H"= -3'2. 



INDEX 



ABERRATION, lateral, 120 
longitudinal, 31, 120 
spherical lens, 89-91 
reflection, 30-33 
refraction, 64-65 
Aplanatism, 31 
Astigmatic pencil, 31-33 
Axial pencil, 18, 68 

BADEN POWELL lens, 104-105 
Brightness, 4-6 

CAMERA lucida, 111 
Cardinal Points, 91-96 

of eye, 99-101 
Caustic, 30-31, 120 
Centrads, 126 
Centre of lens, 82-84 
Centric pencils, 18, 68, 83 
Circle of least aberration, 120 
Circle of least confusion, 117-120 
Coddington lens, 86-87 
Conjugate focus, 19-20 
Conversion of formulae, 65-67 
Critical angle, 37 

DECENTRED lens, 124-126 
Dense medium, 34 
Deviation, prism, 43-44 

prismosphere, 124 

refraction, 36 

rotating mirror, 14-15 
Dioptre, 80-81 

ECCENTRIC pencils, 30-33, 64-65, 

109-110, 111-117 
Eye, 99-102 

FOCAL lines reflection, 30-33, 109- 

110 

refraction, lens, 121-123 
plane surface, 111-113 
plate, 113-114 

spherical surface, 64-65, 114- 
116 



GAUSS, 92 

Graphic methods, for image, 29-30, 

61-64, 80 
for cardinal points, 105-107 

HUYGENIAN eyepiece, 96-97 

ILLUMINATION, 1-4 
Image, lens, 69-77 

plane reflection, 13-17 

plane refraction, 38-40 

prism, 45-46 

real and virtual, 13 

spherical mirror, 18-30 

spherical surface, 50-64 

KALEIDOSCOPE, 15 

LAWS of illumination, 1-4 

propagation, 1-2 

reflection, 11 

refraction, 34-36 
Lenses, 68-107 
Light sense, 3 

MAGNIFICATION, 77-80i 
Mirrors, plane, 11-14 

concave, 18-21 

convex, 22-24 

NODAL points, 84, 87-89, 100-102 

OBLIQUE pencils, lens, 121-123 

plane surface, 111-114 

spherical surface, 30-33, 109-110, 

114-117 
Opera glass, 98 
Optical centre, 82-84 

PHOTOMETER, 3 
Pinholes, 7-8 
Power, 80-81 
Principal points, 91-96 
Prism, 41-47 
Prism dioptres, 124-125 
Prismosphere, 124 



131 



132 INDEX 



REFLECTION plane, 11-15 

repeated, 15-17 

spherical surface, 18-33 
Refraction, plane, 34-47 
Refractive index, 35 
Rotation of mirrors, 14-15 

of prisms, 46-47 

SHADOWS, 8-9 

Signs, 21-22, 49 

Size of image, 26-27, 59-61, 73-77 

Snell, 34 

Sphere, 84-86 



Stanhope lens, 87 

TELESCOPE, penetrating power, 6 
Total reflection, 37 
Tscherning, 99 

ULTEAMICROSCOPE, 6 

Umbra, 9 

Unit planes, 100 

VISUAL angle, 6-7, 77, 102 
WOLLASTON prism, 111 



THE END 



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